physics 6 (sci 33202)

Demonstration School of

Suan Sunandha Rajabhat University

Physics 6 (SCI 33202)

Mathayom 6/1

2nd semester of academic year 2021

Ariyaphol Jiwalak

Preface

The physics book that you are reading now is my attempt to refine the content from the standardized textbooks to suit Thai students. However, there are still some mistakes. And I welcome suggestions from students to improve the books for students in the next generation.

I would like students to realize that physics is a subject that provides a basic understanding of various fields of science. Physics can explain the nature and teaches us to think systematically and logically. Understanding physics is very useful for understanding things that occurs around us. I wrote this book, not for anyone to become a physicist. But I wrote for everyone who studies this course to have some basic physics knowledge and have good experience and good memories in physics.

Ariyaphol Jiwalak 31 March 2021

Content

Page

Chapter 1: Atomic physics 1 1. Thomson’s experiment 2 2. Millikan’s oil drop experiment 5 3. Rutherford’s experiment 7 4. Atomic spectra 9

- Hydrogen 10 5. Blackbody radiation 11

- Plank’s hypothesis 12 6. Bohr’s atomic theory 13

- Radius of electron’s orbit 15 - Total energy of atom 15 - Speed of electron 17

7. Frank-Hertz experiment and X-rays production 22 - Frank-Hertz experiment 22 - X-rays production 23

8. Photoelectric effect 25 9. Compton effect 31 10. Wave-Particle duality 32

- de Broglie’s hypothesis 32 11. Quantum mechanics 37

- Uncertainty principle 37 - Quantum mechanical model of atom 37 - Schrodinger equation 37

Chapter 2: Nuclear physics 38

1. Structure, Mass, and Size of nuclei 39 - Structure 39 - Mass 40 - Size 40

Page

Chapter 2: Nuclear physics (cont.)

2. Nuclear stability 42 - Nuclear force 42 - Binding energy 43

3. Radioactivity 46 - Rays emitted by radioactive elements 46 - Radioactive decay 47

Alpha decay 47 Beta decay 47 Gamma ray emission 47

- Activities 48 - Half-life 51 - Radioactive dice experiment 57

4. Nuclear reaction and Nuclear energy 58 - Fission and Fusion 62

Physics 6 (SCI 33202) Chapter 1: Atomic Physics ___________________________________________________________________________________________

Ariyaphol Jiwalak | Demonstration School of Suan Sunandha Rajabhat University © 2021 1

7สียงแล

1. Thomson’s experiment 2. Millikan’s oil drop experiment 3. Rutherford’s experiment 4. Atomic spectra

- Hydrogen 5. Blackbody radiation

- Plank’s hypothesis 6. Bohr’s atomic theory

- Radius of electron’s orbit - Total energy of atom

Hydrogen energy levels - Speed of electron

7. Frank-Hertz experiment and X-rays production - Frank-Hertz experiment - X-rays production

Continuous X-rays Characteristic X-rays

8. Photoelectric effect 9. Compton effect 10. Wave-Particle duality

- de Broglie’s hypothesis 11. Quantum mechanics

- Uncertainty principle - Quantum mechanical model of atom - Schrodinger equation

Chapter 1: Atomic Physics



DALTON’S ATOMIC MODEL .

- All matter is made of extremely small particles called atoms ‘atoms’, which are indivisible. - All atoms of an element were identical, different elements had atoms of differing size and mass.

1. THOMSON’S EXPERIMENT .

Thomson used a highly evacuated glass container called ‘cathode ray tube’ to measure the ratio of charge to mass for the electron.

In the cathode ray tube, cathode rays from the hot ‘cathode’ are accelerated and formed into a beam by a high potential difference. They pass the ‘anode’ and and strike the screen at the end of the tube, which is coated with a material that fluoresces (glows) at the point of impact.

The cathode rays drift into a region of perpendicular electric field. The electric field produces a defected beam that is deviated to the top of the screen. Thomson concluded that the cathode rays are __________ charged particle beam.

The cathode rays (which has a negative charge) drift into a region of perpendicular magnetic field.

The magnetic field produces a defected beam that is deviated to the __________ of the screen.

qvB = mv2

r

q

m =

v

Br

C A

+ + + + + +

− − − − − −

A C

C A

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The cathode rays drift into a region of perpendicular electric and magnetic fields. The magnitudes of the two fields are adjusted to produce an undeflected beam. The magnitude of the electric force is equal to the magnetic force.

FE = FB

qE = qvB

v = E

B

q

m =

v

Br

q

m =

E

B2r

Therefore, the charge-to-mass ratio for cathode rays: q

m = 1.76 × 1011 C/kg

Thomson repeated the experiment by changing the metal of cathode and found that the charge-

to-mass ratio for cathode rays is always approximately the same. He proposed that the cathode rays that emitted from all metals were the same particles, which were later called ‘electron’.

Thomson's experiments show that negative electrodes made of any metals can provide electrons.

He suggested that “Atoms that were once understood to be indivisible can be further subdivided, and electrons are one component of every atom”.

Electronvolt (eV)

An electronvolt is defined as the amount of kinetic energy gained (or lost) by a single electron accelerating from rest through an electric potential difference of one volt in vacuum.

The work done by the external force on a point charge

W = q∆V

= (1.6 × 10-19 C)(1 V)

= 1.6 × 10-19 Jx

x1 eV = 1.6 × 10-19 Jx

+ + + + + +

− − − − − −

A C

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Example 1.1: You set out to reproduce Thomson’s experiment e/m with an accelerating potential of

150 V and a deflecting electric field of magnitude 6.0 × 106 N/C

a) How fast do the electrons move?

b) What magnetic-field magnitude will yield zero beam deflection?

c) With this magnetic field, how will the electron beam behave if you increase the

accelerating potential above 150 V?



2. MILLIKAN’S OIL DROP EXPERIMENT .

Millikan succeeded in measuring the charge of the electron precisely. In his experiment, oil was sprayed in very fine drops into the space between two parallel horizontal plates separated by a distance d. A potential difference ∆V is maintained between the parallel plates, causing a uniform electric field between them. Some of the oil drops acquire a negative charge and experience an electric force.

By adjusting the potential difference to keep a given drop at rest, Millikan determined the charge on that drop, provided its radius is known. The magnitude of the electric force is equal to the magnetic force.

FE = FG

qE = mg

q (∆V

d) = (ρV)g

where q = charge of oil (NOT charge of electron!)

m = mass of oil (NOT mass of electron!)

Millikan repeated the experiment by changing the oil and found that the charge of oil drop is always equal to an integral multiple of 1.60 × 10-19 C. He proposed that this was the magnitude of the negative charge of a electron (e = 1.60 × 10-19 C) and assume that the negative charge of oil drop is due to an excess electrons.

The charge of the electron, together with the value of its charge-to-mass ratio from Thomson’s experiment, enables us to determine the mass of the electron

e

me = 1.76 × 1011

1.60 × 10-19

me= 1.76 × 1011

me = 9.1 × 10-31 kg

Therefore, the mass of the electron is 9.1 × 10-31 kg.

_ _ _ _

mg

qE

+ + + + + + + + + + +

+ + + +

− − − − − − − − − − − − −



Thomson’s atomic model (Plum pudding model) The discovery of electrons is not consistent with Dolton's atomic model, physicists then proposed

a new model of atoms that electrons are a component of atoms. Thomson suggested “an atom consisted of a sphere of positively charged material in which were

embedded negatively charged electrons. The atom as a whole would be electrically neutral.”

Failure of Thomson’s atomic model Thomson's atomic model could not explain Rutherford's experiments which will be discussed in

the next section.

+

+

+

+

_

_ _

_ _

+ + _



3. RUTHERFORD’S EXPERIMENT .

In Rutherford’s experiment, a beam of positively charged alpha particles (helium nuclei, He24 ) was

projected into a thin gold foil. Most of the particles passed through the foil as if it were empty space. Many of the particles deflected from their original direction of travel were scattered through large angles. Some particles were even deflected backward, completely reversing their direction of travel!

Such large deflections were not expected on the basis of Thomson’s model. According to that model, the positive charge of an atom in the foil is spread out over the entire atom that there is no concentration of positive charge strong enough to cause any large-angle deflections of the positively charged alpha particles. Furthermore, the electrons are so much less massive than the alpha particles that they would not cause large-angle scattering either.

In the Thomson model, the positive charge and the negative electrons are distributed through the whole atom. Hence the electric force on an alpha particle that enters the atom should be quite weak. The maximum deflection to be expected is then only a few degrees.

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Rutherford’s atomic model Rutherford explained his results by developing a new atomic model, one that assumed “the

positive charge in the atom was concentrated in a region called the ‘nucleus’ of the atom that was small relative to the size of the atom. Any electrons belonging to the atom were assumed to be in the relatively large volume outside the nucleus”.

Then the force repelling the alpha particle would be much larger, and the large-angle scattering

could occur. Failure of Rutherford’s atomic model

1. What prevented the electrons from losing the energy due to the electromagnetic radiation? The electrons are described by the particle in uniform circular motion model; they have

a centripetal acceleration. According to Maxwell’s theory of electromagnetism, accelerated charges should radiate electromagnetic waves. As energy leaves the system, the radius of the electron’s orbit steadily decreases. Therefore, the electron moves closer to the nucleus. This process leads to an ultimate collapse of the atom as the electron plunges into the nucleus.

2. How are many electrons arranged in an atom? 3. Why are many positive charges concentrated in the nucleus despite the positive charges positive

charges repel on each other? Example 3.1: An alpha particle (charge 2e) is aimed directly at a gold nucleus (charge 79e). What

minimum initial kinetic energy must the alpha particle have to approach within 5.0 × 10-14

m of the center of the gold nucleus before reversing direction? Assume that the gold

nucleus, which has about 50 times the mass of an alpha particle, remains at rest.

+ + + +

_

_ _

_ _ +



4. ATOMIC SPECTRA .

Hot solid. → Continuous spectrum

Hot gas. → Emission spectrum

Hot filament + Cold gas. → Absorption spectrum

Different hot gases emit different kinds of emission line spectrum. Hence scientist can use atomic spectra to identify elements and compounds.

Hot solid

Hot gas

Hot solid + Cold gas

Line spectrum

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Hydrogen.

The spectral lines of the hydrogen atoms has an orderly wavelength. The complete set of lines is called the ‘series’. The wavelengths of these lines can be described by the following equation.

1

λ = R

H(

1

22 -

1

n2) ; n = 3, 4, 5, …

where RH = Rydberg constant (1.097 × 107 m-1)

Other lines in the spectrum of hydrogen were found following Balmer’s discovery. Series Equation Electromagnetic spectrum

Lyman 1

λ = R

H(

1

12 -

1

n2) ; n = 2, 3, 4, … Ultraviolet

Balmer 1

λ = R

H(

1

22 -

1

n2) ; n = 3, 4, 5, … Visible light

Paschen 1

λ = R

H(

1

32 -

1

n2) ; n = 4, 5, 6, …

Brackett 1

λ = R

H(

1

42 -

1

n2) ; n = 5, 6, 7, … Infrared

Pfund 1

λ = R

H(

1

52 -

1

n2) ; n = 6, 7, 8, …

No theoretical basis existed for these equations at the time; they simply worked, but nobody knew

why does the wavelengths of these lines conform to these equations, and why the wavelengths is related to the integer n.

Balmer series of spectral lines for atomic hydrogen



5. BLACK BODY RADIATION .

A black body is an ideal system that absorbs all radiation incident on it; there is no reflection at all from its surface. A good approximation to a blackbody is a hollow box with a small aperture in one wall. Any radiation incident on the hole from outside the cavity enters the hole and is reflected a number of times on the interior walls of the cavity; hence, the hole acts as a perfect absorber.

The nature of the radiation leaving the cavity through the hole depends only on the temperature of the cavity walls and not on the material of which the walls are made.

The following consistent experimental findings were seen as especially significant: - The total intensity of the radiation from blackbody increases with temperature. - The intensity is distributed over all wavelengths, but the shape of the distribution function

is the same for all temperatures; that is the peak of the wavelength distribution shifts to shorter wavelengths as the temperature increases.



Planck’s hypothesis.

In order to explain the wavelength distribution of radiation from a blackbody Planck proposed the assumption that “the energy of electromagnetic wave of frequency f that a black body emits (or absorbs) can have only certain discrete values, which is equal to an integral multiple of hf. The discrete energy is called ‘quantum of energy’.” That is the energy of the electromagnetic radiation is discontinuous.

The energy E of the electromagnetic wave of frequency f

E = nhf

where n = quantum number (1, 2, 3, …) h = Planck’s constant (6.6 × 10-34 J s)



6. BOHR’S ATOMIC THEORY .

Bohr combined ideas from Planck’s hypothesis (the positive charge was concentrated in the nucleus of the atom) and Rutherford’s atomic model (the energy of the electromagnetic radiation is discontinuous, and the electron moves in circular orbits around the nucleus) to arrive at a theory of hydrogen atom. The Bohr’s atomic theory as it applies to the hydrogen atom has two assumtions: 1. The electron moves in circular orbits around the nucleus without releasing the energy in the form of

electromagnetic waves. In this stable orbit the electron has a constant angular momentum L which must be an integral multiple of h̅.

Therefore, the magnitude of the angular momentum of an electron of mass m traveling around the nucleus with linear speed v and radius r.

L = nh̅

mvr = nh̅

where n = 1, 2, 3, …

h̅ = h

2π = 1.05 × 10-34 J s

2. An electron can make a transition from one energy level to another level by emitting (or absorbing) an

electromagnetic waves of frequency f with energy equal to the energy difference between the initial and final levels.

Ei - Ef = hf

Consider the hydrogen atom consisting of one proton and one electron.

The three main equations

1. L = mvr = nh̅

2. ΣFc = mv2

r →

k(e)(e)

r2 =

mv2

r

3. E total of the atom = Ep atom + Ek electron = k(+e)(-e)

r +

1

2mv2

p+

e- of mass m

vሬԦ

Fc r



Derivation rn = a0n2

En = -13.60

n2 eV

vn = 2.18 × 106

n



Radius of electron’s orbit.

rn = a0n2

where a0 = 0.53 × 10-10 m is called ‘Bohr radius’

→ Radius of the first orbit r1 = 12 a0 = 1 a0

→ Radius of the second orbit r2 = 22 a0 = 4 a0

→ Radius of the third orbit r3 = 32 a0 = 9 a0 ...

Total energy of atom.

En = -13.60

n2 eV

The energies of atoms are quantized: They can have only certain definite values. Bohr assumed that the nucleus was not moving (no kinetic energy), so the total energy of the atom

is the total energy of the electrons.

• Hydrogen energy levels

- The lowest level or the ‘ground state’ has n = 1 and minimum energy (most stable). - When the electrons are given energy (stimulated), the electrons will make a transition to a higher

level called the ‘excited state'. The electrons in the excited state will return to the ground state and will release energy in the form of electromagnetic waves of wavelength λ.

λ (nm) = 1240

Ei - Ef (eV)

- The uppermost level has n = ∞ and E = 0.

E6 = _____ eV

E1 = _____ eV ground state

E2 = _____ eV 1st excited state

E3 = _____ eV 2nd excited state

E4 = _____ eV

E5 = _____ eV

E∞ = _____ eV

e-

λ = 1240

10.2 eV = 122 nm

10.2 eV



Derivation λ (nm) = 1240

Ei - Ef (eV)

Bohr's atomic theory can be used to describe the spectrum of hydrogen. “The emission

spectrum is emitted due to an electron making a transition from a high energy level to a lower energy level by releasing energy in the form of electromagnetic waves”.



Derivation Spectral lines of the hydrogen atoms 1

λ = R

H(

1

nf2

- 1

ni2)

Speed of electron.

vn = 2.18 × 106

n

Failure of Bohr’s atomic theory

The Bohr’s atomic theory can only explain hydrogen atoms consisting of one proton and one electron, it could not be used to describe an atom with two or more electrons.



Example 6.1: For a hydrogen atom, calculate the energies of states corresponding to n = 2.

Example 6.2: What is the radius of the electron orbit for a hydrogen atom for which n = 5?

Example 6.3: How fast is the electron moving in a hydrogen atom for which n = 5?



Example 6.4: Find the kinetic, potential, and total energies of the hydrogen atom in the first excited

level, and find the wavelength of the photon emitted in a transition from that level to the

ground level.

Example 6.5: The electron in a hydrogen atom makes a transition from a higher energy level to the

ground level (n = 1). Find the wavelength and frequency of the emitted photon if the

higher level is n = 2.

Example 6.6: Suppose the hydrogen atom is initially in the higher level corresponding to n = 5. What is

the wavelength of the photon emitted when the atom drops from n = 5 to n = 1?



Example 6.7: A hypothetical atom has energy levels at 0.00 eV (the ground level), 1.00 eV, and 3.00 eV.

a) What are the frequencies and wavelengths of the spectral lines this atom can emit

when excited?

b) What wavelengths can this atom absorb if it is in its ground level?



Example 6.8: Consider the energy level diagram for a particular atom shown below.

Energy above ground state

E4 = 7 eV

E3 = 6 eV

E2 = 4 eV

E1 = 0 eV a) An electron begins in the ground state of this atom. How much energy must be

absorbed by the electron to reach the fourth energy level?

b) How many possible ways can this atom emit a photon if the electron starts in the

fourth energy level?

c) The electron drops from E4 to E2 and emits a photon, then drops from E2 to E1 and

emits a second photon. Which of these emitted photons has the higher frequency?



7. FRANCK-HERTZ EXPERIMENT AND X-RAY PRODUCTION: Are energy levels real? .

Franck-Hertz experiment.

Franck and Hertz studied the motion of electrons through mercury vapor under the action of an electric field. They found that when the electron kinetic energy was 4 . 9 eV or greater, the vapor emitted ultraviolet light of wavelength 253 nm.

Suppose mercury atoms have an excited energy level 4.9 eV above the ground level. An atom can be raised to this level by collision with an electron; it later decays back to the ground level by emitting a photon.

From the photon formula, the wavelength of the photon should be λ = 1240

4.9 eV = 253 nm. This is

equal to the wavelength that Franck and Hertz measured.

4.9 eV

Mercury atom

E2 (1st excited state)

E1 (ground state) e-

e-

4.9 eV

λ = 1240

4.9 eV = 253 nm

Voltage (V)

Curr

ent

(mA

)

ไส้หลอด

(filament) grid

collector

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X-rays production.

When the electrode is heat to a very high temperature, it releases electrons. The electrons are then accelerated toward the metal target. X-rays are produced when high-energy electrons strike a metal target.

• Continuous X-rays The continuous x-rays are the result of the slowing down of high-energy electrons as they

strike the metal target. It may take several interactions with the atoms of the target before the electron gives up all its kinetic energy. The amount of kinetic energy given up in any interaction can vary from zero up to the entire kinetic energy of the electron. Therefore, the wavelength of radiation from these interactions lies in a continuous range from some minimum value up to infinity.

The most energetic x-ray (shortest wavelength) is produced if the electron is braked to a stop all at once when it hits the metal target, so that all of its kinetic energy goes to produce x-rays.

Example 7.1: Electrons in an x-ray tube accelerate through a potential difference of 10.0 kV before

striking a target. If an electron produces one photon on impact with the target, what is the

minimum wavelength of the resulting x rays? Find the answer by expressing energies in

both SI units and electron volts.



• Characteristic X-rays The characteristic x-rays occurs when a high-energy electron collides with a target atom. The

electron must have sufficient energy to remove an inner-shell electron from the atom. The vacancy created in the shell is filled when an electron in a higher level drops down into the level containing the vacancy, emitting an x-ray in the process. Each target atom has definite energy, so the emitted x-rays have definite wavelengths; they form what is called ‘characteristic x-ray’ for each target atom.



8. PHOTOELECTRIC EFFECT: Light absorbed as photons .

Hertz’s experiments showed that light incident on certain metallic surfaces causes electrons to be emitted from those surfaces. The emitted electrons are called ‘photoelectrons’.

When the cathode C is illuminated by light, an electron is emitted and moved to the anode A. The photoelectrons is detected by reading a current on the ammeter.

We can determine the number of photoelectrons by making the positive potential of the anode relative to the cathode, so that the anode A attract all of the electrons from the anode C

When the light intensity increases, the current flowing through the ammeter increases Therefore, it can be concluded that the number of photoelectrons will increase as the light intensity increases

We cannot use the properties of waves to explain the results of this experiment. Because if the

light is a wave - Since the high intensity light has high energy, the electrons with high kinetic energy should be

ejected from the metal at any incident light frequency, as long as the light intensity is high enough.

- Electrons should be ejected from the metal at incident light with a frequency lower than the threshold frequency, as long as the time is long enough, because the electrons will gradually absorbed more and more energy from the light it acquires enough energy to escape from the metal.

- The time interval is required for the electron to absorb the incident radiation before it acquires enough energy to escape from the metal. Electrons should not be emitted from the surface of the metal almost instantaneously.

C A

A

C A

A

C A

A



Einstein made the postulate based on Planck's hypothesis that a beam of light consists of small packages of energy called ‘photons’. Each photon has an energy E given by equation, E = hf. In Einstein’s picture, an individual photon arriving at the surface is absorbed by a single electron. The electron gets all of the photon’s energy or none at all. The electron can escape from the surface only if the energy it acquires is greater than the ‘work function (w)’. The work function represents the energy with which an electron is bound in the metal, which is characteristic of the metal.

Einstein applied conservation of energy to find that the maximum kinetic energy for an emitted electron is the energy gained from a photon minus the work function:

Ek, max = hf - w

This equation is called photoelectric equation

slope = _______

y-intercept = _______

- Photoelectrons will NOT be ejected if the energy of photon hf is less than the work function

w. - Photoelectrons will be ejected with zero kinetic energy only if the energy of photon hf is equal

to the work function w. The frequency of the light equals to the threshold frequency: 0 = hf0 - w

w = hf0

- Photoelectrons will be ejected with non-zero kinetic energy only if the energy of photon hf is more than the work function w.

Greater intensity means a greater number of photons per second absorbed, and thus a greater number of electrons emitted per second, but the photoelectrons have the same frequency.

Greater frequency means a greater energy of photons, and thus a greater energy of electrons emitted, but the number of photoelectrons is not change.

Ek

f



We can determine the maximum kinetic energy of the emitted electrons by making the potential of the anode relative to the cathode, just negative enough so that the current stops.

Using the law of conservation of energy, we have Ek, max = e∆Vs

where ∆Vs = stopping potential (V)

The stopping potential is the potential necessary to stop any electron just before reaching the anode and the current is zero.

This equation states that electrons with high kinetic energy require high stopping potential.

Experiments showed that - The stopping potential does not depend on intensity; the stopping potential does not change

as the light frequency increases. - The stopping potential does depend on frequency; the stopping potential increases as the

light frequency increases. No electrons are emitted if the incident light frequency falls below some ‘threshold frequency ( f0 )’, whose value is characteristic of the material being illuminated.

I

∆V

C A

A



(1) Ek = hf - w

Ek

e =

hf

e -

w

e

Ek (eV) = hf (eV) - w (eV)

Ek (eV) = 1240

λnm - w (eV)

(2) Ek = hf - w

e∆Vs = hf - w

∆Vs = hf

e -

w

e

∆Vs = hf (eV) - w (eV)

slope = _______

y-intercept = _______

∆Vs

f



Example 8.1: A sodium surface is illuminated with light having a wavelength of 300 nm. The work

function for sodium metal is 2.46 eV.

a) Find the maximum kinetic energy of the ejected photoelectrons.

b) Find the cutoff wavelength λc for sodium.

Example 8.2: While conducting a photoelectric-effect experiment with light of a certain frequency, you

find that a reverse potential difference of 1.25 V is required to reduce the current to zero.

Find

a) the maximum kinetic energy

b) the maximum speed of the emitted photoelectrons



Example 8.3: For a particular cathode material in a photoelectric-effect experiment, you measure

stopping potentials as a function of frequency as in Fig. below. Determine the work

function W for this material and the implied value of Planck’s constant h.

Example 8.4: The metal sodium has a threshold frequency that corresponds to yellow light, describe

what would happen in the following instances:

a) We shine orange light on the sodium metal.

b) We shine very bright red light on the sodium metal.

c) We shine blue light on the sodium metal.

d) We shine very bright blue light on the sodium metal.

s



9. COMPTON EFFECT: X-rays scattered as photons .

Compton aimed a beam of x rays of wavelength λ0 at an electron in graphite target and measured

the wavelength of the radiation λ' scattered from the target. He discovered that some of the scattered

radiation has the same wavelength as the incident radiation (λ' = λ0), and some of the scattered radiation

has longer wavelength than the incident radiation (λ' > λ0).

before collision after collision

We cannot use the wave properties of X-rays to explain the results of this experiment. But we have to use Einstein's photons theory; that is the collision between photon of x-rays and electron in a graphite is similar to a collision of particles, which can be explained by the law of energy conservation and the law of conservation of momentum. The scattering X-ray has the same wavelength as the incident X-ray because it is an elastic collision, and the scattering X-ray has more wavelengths (decrease in energy) than the incident X-ray because it is an inelastic collision. Therefore, The Compton effect supports Einstein's theory that the electromagnetic waves (X-rays) have particle properties.

λ0 m θ

∅

m

λ'



10. WAVES-PARTICLES DUALITY .

de Broglie’s hypothesis.

If an electromagnetic waves act like a particle such as the photoelectric effect and the Compton effect, it should have a momentum, which is a particle’s property.

de Broglie proposed that the momentum P of a wave with wavelength λ can be expressed as

P = h

λ

Derivation P =h

λ

de Broglie postulated that because waves have both wave and particle characteristics such as the photoelectric effect and the Compton effect, perhaps all forms of particles such as electrons, protons, neutrons, atoms, molecules, stones, or balls have both properties.

If a particle acts like a wave, it should have a wavelength. De Broglie postulated that a particle with mass m, moving with speed v should have a wavelength λ related to its momentum. The wavelength of a particle is then

λ = h

mv

The wavelength of the particle or matter waves is called ‘de Broglie wavelength’. From de Broglie’s hypothesis, it can be concluded that “Waves may in some situations behave like

particles, and particles may in some situations behave like waves”. This is called ‘wave-particle duality’.



Rather than visualizing the orbiting electron as a particle moving around the nucleus in a circular path, think of it as a sinusoidal standing wave with wavelength λ that extends around the circle. An electrons in Bohr’s orbits radiate no energy because the wave come out even and join onto itself smoothly. The circumference of this circle must include some whole number of wavelengths.

Instead of going through Bohr’s argument to justify equation mvr = nh̅, we can use de Broglie’s picture of electron waves.

2πr = nλ

2πr = n (h

mv)

mvr = n (h

2π)

mvr = nh̅

https://www.google.co.th/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&ved=0ahUKEwjbp4KBgPDWAhVMv5QKHTWPC5EQygQILDAA&url=https%3A%2F%2Fen.wikipedia.org%2Fwiki%2FPi%23Adoption_of_the_symbol_.CF.80&usg=AOvVaw3pdIaiiVU2BbzSyYCBmJp3





Example 10.1: A laser pointer with a power output of 5.00 mW emits red light (λ = 650 nm).

a) What is the magnitude of the momentum of each photon?

b) How many photons does the laser pointer emit each second?



Example 10.2: Calculate the de Broglie wavelength for an electron (me = 9.11 × 10-31 kg) moving at 1.00

× 107 m/s. Example 10.3: A rock of mass 50 g is thrown with a speed of 40 m/s. What is its de Broglie wavelength? Example 10.4: If an electron, a proton, and an alpha particle (a helium nucleus) are all moving at the

same speed, which will have the longest de Broglie wavelength? Which will have the

shortest?



Example 10.5: Find the speed and kinetic energy of a neutron (m = 1.675 × 10-27 kg) with de Broglie wavelength λ = 0.200 nm.

Example 10.6: What accelerating voltage is needed to produce electrons with wavelength 0.010 nm?



11. QUANTUM MECHANICS .

Uncertainty principle.

Heisenberg postulated that “it is physically impossible to measure simultaneously the exact position and exact momentum of a particle”. The product of the uncertainties in the x-component of

position and momentum can never be smaller than h̅.

(Δx)(ΔPx) ≥ h̅

where ∆x = uncertainty in the x-component of position

∆Px = uncertainty in the x-component of momentum

Example 11.1: An electron is confined within a region of width 5.000 × 10-11 m. Estimate the minimum

uncertainty in the x-component of the electron’s momentum.

Quantum mechanical model of atom.

According to the uncertainty principle, we cannot know exactly where an electron is at any given time or how they move around the nucleus of the atom, but we can only know the opportunity to find electrons at any positions in the atom. Image of the electrons in atoms are similar to fog clusters that envelop the nucleus. The position that the fog has high intensity indicates that the area is more likely to find electrons than other areas. In addition, the quantum mechanics can be used to describe an atom with two or more electrons. Schrodinger equation.

If the particles may in some situations behave like waves according to De Broglie's hypothesis, the equations of motion of the electrons should be similar to the wave equation, which will be studied at a higher level. The result of solving Schrödinger's equation indicates the discontinuity of energy and angular momentum of the electron, which corresponds to Bohh's atomic theory.

https://en.wiktionary.org/wiki/%CE%94

https://en.wiktionary.org/wiki/%CE%94

Physics 6 (SCI 33202) Chapter 2: Nuclear and Particle Physics ___________________________________________________________________________________________


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1. Structure, Mass, and Size of nuclei - Structure

Proton-Neutron hypothesis - Mass - Size

2. Nuclear stability - Nuclear force - Binding energy

3. Radioactivity - Rays emitted by radioactive elements - Radioactive decay

Alpha decay Beta decay Gamma ray emission

- Activity - Half-life - Radioactive dice experiment

4. Nuclear reaction and nuclear energy - Fission and Fusion

Fission and Chain reactions Fusion

Chapter 2: Nuclear and Particle Physics



1. STRUCTURE, MASS, AND SIZE OF NUCLEI .

Structure of nuclei.

• Proton-Neutron hypothesis “The nucleus consists of protons (which has a positive electric charge) and neutrons (which

are electrically neutral). These particles, which are components of nucleus, are called ‘nucleon’.”

Isotopes, Isotones, and Isobars - Isotopes: same number of protons but a different numbers of neutrons - Isotones: same number of neutrons but a different numbers of protons - Isobars: same mass number

The symbol of the element

XZA

where A = mass number

(The sum of the number of protons and the number of neutrons (p+ + n)) Z = atomic number (The number of protons in the nucleus (p+))

such as proton H11

deuteron H12

tritron H13

neutron n01

electron e-10

Example 1.1: Determine the number of protons, neutrons, and electrons in a neutral atom of iron Fe2656 .



Mass of nuclei.

C-12 6.0221409 × 1023 atoms → mass 12.000000 g

C-12 1 atom → mass 12

6.0221409 × 1023 g

1/12 of C-12 1 atom → mass 1

12(

12

6.0221409 × 1023) g

0.1660540 × 10-23 g

1.660540 × 10-27 kg

∴ C-12 1 atom → mass 12.000000 u

Size of nuclei.

The volume V of a nucleus is proportional to the nucleon number A V ∝ A

4

3πR3 ∝ A

R ∝ A1

3 Therefore, the radii of most nuclei are represented by the equation

R = r0√A3

where r0 = 1.2 × 10-15 m

Particle Mass

(kg) (u)

1/12 of the mass of an atom of C-12 1.660540 × 10-27 1.000000

proton 1.672623 × 10-27 1.007276

neutron 1.674629 × 10-27 1.008665

electron 9.109390 × 10-31 0.000549



Example 1.2: The most common kind of iron nucleus has mass number A = 56. Find the radius,

approximate mass, and approximate density of the nucleus.

Example 1.3: Consider a nucleus of mass number A, containing protons and neutrons, each with mass

approximately equal to m.

a) Find an approximate expression for the mass of the nucleus.

b) Find an expression for the volume of this nucleus in terms of A.

c) Find a numerical value for the density of this nucleus.



2. NUCLEAR STABILITY .

Nuclear force.

You might expect that the very large repulsive Coulomb forces between the closepacked protons in a nucleus should cause the nucleus to fly apart. Because that does not happen, there must be a counteracting attractive force. The nuclear force is an attractive force that acts between all nuclear particles. The protons attract each other by means of the nuclear force, and, at the same time, they repel each other through the Coulomb force. The nuclear force also acts between pairs of neutrons and between neutrons and protons. The nuclear force dominates the Coulomb repulsive force within the nucleus, so stable nuclei can exist.

The stable nuclei are represented by the black dots, which lie in a narrow range called the line of stability. Notice that the light stable nuclei contain an equal number of protons and neutrons. Also notice that in heavy stable nuclei, the number of neutrons exceeds the number of protons, the line of stability deviates upward from the line representing N = Z. This deviation can be understood by recognizing that as the number of protons increases, the strength of the Coulomb force increases, which tends to break the nucleus apart. As a result, more neutrons are needed to keep the nucleus stable because neutrons experience only the attractive gravitational force.



Binding energy (B.E.).

The binding energy is the energy that must be added to a nucleus to break it apart into its components or the magnitude of the energy by which the nucleons are bound together.

The binding energy of any nucleus can be calculated by using the conservation of energy and the Einstein mass–energy equivalence relationship (E = mc2) Note: Mass 1 u can convert into energy 931 MeV

Derivation Mass 1 u can convert into energy 931 MeV



Example 2.1: Calculate the binding energy of the C612 .

Mass of nucleus = 12.00000000 u – 6 (0.00054858 u)

= 11.99670852 u Mass of nucleon = 6 (1.007276 u) + 6 (1.008664 u)

= 12.09564 u

We see that the mass of a nucleus is always less than the total mass of its nucleons, called the ‘mass defect (∆m)’. The lost mass becomes the binding energy that holds protons and neutrons together in the nucleus.

Mass defect = 12.09564 u – 11.99670852 u = 0.09893148 u Binding energy = 0.09893148 × 931 MeV = 92.16 MeV

Example 2.2: Calculate the binding energy of the Li37 .

(The nucleus of lithium has a mass of 7.016005 u.)

Note: When the number of nucleons in nuclei increases, the binding energy increases continuously with

mass number.



Stop to think Which nucleus is more stable

between U92236 which has binding energy 1,796 MeV

and Ni2858 which has binding energy 508 MeV?

Answer: An important measure of how tightly a nucleus is bound is the ‘binding energy per nucleon’.

Example 2.3: Find the mass defect, the total binding energy, and the binding energy per nucleon of Ni28

62 ,

which has the highest binding energy per nucleon of all nuclides. The neutral atomic mass

of Ni2862 is 61.928345 u.



3. RADIOACTIVITY .

Radioactive element Radioactive elements are elements that can emit rays by themselves.

Radioactivity Radioactivity is a process that the unstable nuclides decay to form other nuclides by emitting

particles and electromagnetic radiation.

Rays emitted by radioactive elements.

• Alpha ray (α, He24 ) is a helium nucleus.

• Beta ray (β, e-10 ) is a high-energy electron.

• Gamma ray (γ) is a high-frequency electromagnetic wave.

Note: Ray is a small group of particles which moves continuously at a high speed as a beam

Mass, Energy: α > β > γ

Velocity, Penetrating power: γ > β > α



Radioactive decay.

• Proton-rich unstable nuclei.

Alpha decay

For example , U92238 → Th + He2

490

234

Stop to think Compare the ratio of neutron to protons between U92238 and Th90

234 Beta-plus or Positron ( e+1

0 ) decay

p11 → n0

1 + e+10 + νe

For example , N714 → C6

14 + e+10 + νe

• Neutron-rich unstable nuclei.

Beta-minus decay

n01 → p

11 + e -1

0 + v̅e

For example , C614 → N7

14 + e-10 + ν̅e

• Nucleus in an excited state.

Gamma ray emission

For example , C* → C 6

12+ γ

6

12



Activities.

No change in physical or chemical environment, such as chemical reactions or heating or increasing pressure, greatly affects the radioactive decays. “The rate of change of the number of radioactive nuclei in a sample or the ‘decay rate (A)’ is proportional to the number of radioactive nuclei present”.

A ∝ N

A = λN

where λ = decay constant (s-1)

Decay constant is the probability of decay per nucleus per second. It has different values for different nuclides. A large value of λ corresponds to rapid decay; a small value corresponds to slower decay.

The decay rate or the ‘activity’ is the number of decays per second. The unit of of activity is decays per second (s-1) or ‘Becquerel’ (Bq) A frequently used unit of activity is the ‘Curie’ (Ci), defined as 1 Ci = 3.7 × 1010 Bq

The radioactive element has an initial number of radioactive nuclei N0 and has a decay constant

λ. Over time t, the number of undecayed nuclei remaning N is

N = N0e-λt

N

t



Derivation N = N0e-λt



Example 3.1: At time t = 0, a radioactive sample contains 3.50 µg of pure C611 , which has a half-life of

20.4 min. a) Determine the number of nuclei in the sample at t = 0. b) What is the initial activity of the sample?

Example 3.2: The isotope Cox57 decays by electron capture to Fex

57 with a halflife of 272 d. The Fex57

nucleus is produced in an excited state, and it almost instantaneously emits gamma rays that we can detect.

a) Find the decay constant for Cox57

b) If the activity of a Cox57 radiation source is now 2.00 µCi, how many Cox

57 nuclei does the source contain?



Half-life (T1

2

).

The half-life is the time required for the number of radioactive nuclei to decrease to one-half the original number.

In general, after n half-lives, the number of undecayed radioactive nuclei remaining is

N = N0

2n

where n = t

T1

2

The relationship between the half-life T1

2

and the decay constant λ

λ = ln 2

T12

This equation states that the half-life T1

2

of an unstable nucleus is inversely proportional to the

decay constant λ.

N

t

T12

T12

T12

T12

T12

N0

32

N0

16

N0

8

N0

4

N0

2

N0



Derivation The relationship between the half-life T1

2

and the decay constant λ λ = ln 2

T12



Example 3.3: Cobalt Co2760 has a half-life of 5 years. If we start with a 100-gram sample of cobalt, how

much cobalt remains after 20 years?

Example 3.4: The isotope carbon-14, C614 , is radioactive and has a half-life of 5,730 years. If you start

with a sample of 1,000 carbon-14 nuclei, how many nuclei will still be undecayed in 25,000 years?



Example 3.5: At time t = 0, a radioactive sample contains 3.50 µg of pure C611 , which has a half-life of

20.4 min. What is the activity of the sample after 8.00 h?

Example 3.6: The isotope Cox57 decays by electron capture to Fex

57 with a halflife of 272 d. The Fex57

nucleus is produced in an excited state, and it almost instantaneously emits gamma rays

that we can detect. If the activity of a Cox57 radiation source is now 2.00 µCi, What will be

the activity after one year?



Example 3.7: A sample of the isotope Ix131 , which has a half-life of 8.04 days, has an activity of 5.0 mCi

at the time of shipment. Upon receipt of the sample at a medical laboratory, the activity is 2.1 mCi. How much time has elapsed between the two measurements?

Example 3.8: A piece of charcoal containing 25.0 g of carbon is found in some ruins of an ancient city.

The sample shows a Cx14 activity R of 250 decays/min. How long has the tree from which this charcoal came

been dead? The ratio of Cx14 to Cx

12 in the carbondyoxide molecules of our atmosphere has a constant value

of approximately 1.3 × 10-12. The Cx14 decays with a half-life of 5730 years.



Example 3.9: Before 1900 the activity per unit mass of atmospheric carbon due to the presence of Cx14

averaged about 0.255 Bq per gram of carbon.

a) What fraction of carbon atoms were Cx14 ?

b) In analyzing an archaeological specimen containing 500 mg of carbon, you observe

174 decays in one hour. What is the age of the specimen, assuming that its activity per unit mass of carbon when it died was that average value of the air?



Radioactive dice experiment.

Comparison of radioactive decay and radioactive dice experiment.

Radioactive decay Radioactive dice experiment

Number of undecayed radioactive nuclei Number of dice remaining

Number of decayed radioactive nuclei Number of dice removed at each throw

Elapsed time Number of throws

Decay constant Probability of one dice being a particular number

Half-life Number of throws that the number of dice to decrease to one-half the original number

Number of dice remaining

Number of throws



4. NUCLEAR REACTION AND NUCLEAR ENERGY .

Nuclear reactions is a process in which a nucleus changes its composition or energy level which may occur spontaneously (such as the decay of unstable nuclei) or by bombarding nucleus with energetic particles.

As a general rule in any nuclear reaction, - the sum of the mass numbers must be the same on both sides of the equation - the sum of the atomic numbers must be the same on both sides of the equation.

Example 4.1: The element tritium H13 is combined with another element to form helium He2

4 and a neutron, along with the release of energy. The equation for this fusion reaction is

H13 + XZ

A → He24 + n0

1 What is the unknown element X?

The difference between the masses before and after the reaction corresponds to the ‘reaction energy’, according to the mass–energy relationship E = mc2.

Example 4.2: N714 + He2

4 → O817 + H1

1

14.003074 u + 4.002604 u 16.999134 u + 1.007825 u 18.005678 u 18.006959 u

The mass increases by = 18.006959 u – 18.005678 u

= 0.001281 u

and the corresponding reaction energy is = 0.001281 u × 931 MeV

= 1.19 MeV (Endoergic)

(The bombarding particle ( He24 ) has a kinetic energy greater than 1.19 MeV.)



Example 4.3: Li37 + H1

1 → He24 + He2

4

7.016005 u + 1.007825 u 4.002604 u + 4.002604 u 8.023830 u 8.005208 u

The mass decreases by = 8.023830 u – 8.005208 u

= 0.018622 u

and the corresponding reaction energy is = 0.018622 × 931 MeV

= 17.34 MeV (Exoergic)

The reaction energy may be in the form of moving particles or in the form of electromagnetic waves.

Conclusions: Mass increases → Endoergic Mass decreases → Exoergic



Example 4.4: Calculate the reaction energy for the reaction He24 + N7

14 → O + H11

817 .

Example 4.5: When a lithium-7 nucleus is bombarded by a proton, two alpha particles ( He24 ) are

produced. Find the reaction energy.



Example 4.6: The Rax226 nucleus undergoes alpha decay according to the following equation.

Ra88226 → Rn + He2

486

222 Calculate the energy for this process.

The masses are 226.025408 u for Rax226

222.017576 u for Rnx222

and 4.002603 u for He24

Example 4.7: Show that the α-emission process Ra88226 → Rn + He2

486

222 is energetically possible, and calculate the kinetic energy of the emitted α particle. The neutral atomic masses are

226.025410 u for Ra88226 , 222.017578 u for Rn86

222 , and 4.002603 u for He24 .



Fission and Fusion.

• Fission Nuclear fission occurs when a heavy nucleus splits into two smaller nuclei. The fission process

often releases a very large amount of energy. The binding energy per nucleon after the reaction is greater than before.

For example, nuclear fission in the nuclear reactors

Nuclear equation: U 92235 + n → Ba56

141

0

1 + Kr + 3 n0

136

92 + 200 MeV

Example 4.8: A fission reaction occurs when uranium He92235 absorbs a slow neutron and then

splits into xenon and strontium, releasing three neutrons and some energy. The equation for this fission reaction is

U92235 + n0

1 → Xe54140 + SrZ

A + 3 n01 + energy

What are Z and A for strontium?

Example 4.9: Calculate the energy released when 1.00 kg of Ux235 fissions, taking the

disintegration energy per event to be 208 MeV.

Example 4.10: What mass of Ux235 must undergo fission each day to provide 3000 MW of thermal

power?



Chain reactions Chain reaction refers to a process in which neutrons released in fission produce an additional

fission in further nucleus.



• Fusion Nuclear fusion occurs when two small nuclei combine to form a heavier nucleus. Fusion

reactions release energy for the same reaction as fission reactions. The binding energy per nucleon after the reaction is greater than before.

For example, nuclear fusion in the sun

Nuclear equations: H11 + H1

1 → H + e+10 + ν

1

2

H12 + H1

1 → He + γ2

3

He23 + He2

3 → He + 2 H11

2

4

Combined equation: 4 H11 → He + 2 e + 2+1

0ν + 2γ + 2

2

46 MeV

Example 4.11: Two deuterons fuse to form a triton (a nucleus of tritium, or Hx3 ) and a proton.

How much energy is liberated?

Physics 6 (SCI 33202) Assignment ___________________________________________________________________________________________


7สียงแล

Problem 1: A photon of green light has a wavelength of 520 nm. Find the photon’s frequency and

energy. Express the energy in both joules and electron volts. Problem 2: A 75-W light source consumes 75 W of electrical power. Assume all this energy goes into

emitted light of wavelength 560 nm. a) Calculate the frequency of the emitted light. b) How many photons per second does the source emit?

Assignment 1: Bohr's atomic theory



Problem 3: In the Bohr theory of the hydrogen atom, an electron moves in a circular orbit about a

proton, where the radius of the orbit is 5.29 × 10-11 m. a) Find the magnitude of the electric force exerted on each particle.

b) Find the magnitude of the gravitational force exerted on each particle.

c) If the electric force causes the centripetal acceleration of the electron, what is the

speed of the electron?

Problem 4: In the Bohr model of the hydrogen atom, an electron moves in a circular path around a

proton. The speed of the electron is approximately 2.20 × 106 m/s. Find

a) the force acting on the electron as it revolves in a circular orbit of radius 0.529 × 10-10 m

b) the centripetal acceleration of the electron.



Problem 5: What is the energy of a photon that, when absorbed by a hydrogen atom, could cause an electronic transition from a) the n = 2 state to the n = 5 state?

b) the n = 4 state to the n = 6 state?

Problem 6: A hydrogen atom initially in its ground level absorbs a photon, which excites the atom to the n = 6 level. Determine the wavelength and frequency of the photon.



Problem 7: The Balmer series for the hydrogen atom corresponds to electronic transitions that terminate in the state with quantum number n = 2 as shown in Figure below.

Consider the photon of longest wavelength corresponding to a transition shown in the figure. Determine a) its energy

b) its wavelength

Consider the spectral line of shortest wavelength corresponding to a transition shown in the figure. Find a) its photon energy

b) its wavelength

c) What is the shortest possible wavelength in the Balmer series?



Problem 8: Calculate the wavelengths of the first three lines in Lyman series and identify the region of the electromagnetic spectrum in which these lines appear.

Problem 9: Find the longest and shortest wavelengths in the Paschen series for hydrogen. In what

region of the electromagnetic spectrum does each series lie??



Problem 10: A monochromatic beam of light is absorbed by a collection of ground-state hydrogen atoms in such a way that six different wavelengths are observed when the hydrogen relaxes back to the ground state. a) What is the wavelength of the incident beam?

b) What is the longest wavelength in the emission spectrum of these atoms?

c) What is the shortest wavelength in the emission spectrum of these atoms?



Problem 11: The energy-level scheme for the hypothetical one-electron element Searsium is shown in Fig. below. The potential energy is taken to be zero for an electron at an infinite distance from the nucleus.

a) How much energy (in electron volts) does it take to ionize an electron from the ground level?

b) An 18-eV photon is absorbed by a Searsium atom in its ground level. As the atom returns to its ground level, what possible energies can the emitted photons have? Assume that there can be transitions between all pairs of levels.

c) What will happen if a photon with an energy of 8 eV strikes a Searsium atom in its ground level?



Problem 12: An atom initially in an energy level with E = -6.52 eV absorbs a photon that has wavelength 860 nm. What is the internal energy of the atom after it absorbs the photon?

An atom initially in an energy level with E = -2.68 eV emits a photon that has wavelength 420 nm. What is the internal energy of the atom after it emits the photon?



Problem 13: An isolated atom of a certain element emits light of wavelength 520 nm when the atom falls from its fifth excited state into its second excited state. The atom emits a photon of wavelength 410 nm when it drops from its sixth excited state into its second excited state. Find the wavelength of the light radiated when the atom makes a transition from its sixth to its fifth excited state.



Problem 14: In a set of experiments on a hypothetical one-electron atom, you measure the wavelengths of the photons emitted from transitions ending in the ground level (n = 1) as shown in the energy-level diagram in Fig below. You also observe that it takes 17.50 eV to ionize this atom.

a) What is the energy of the atom in each of the levels (n = 1, n = 2, etc.) shown in the figure?

b) If an electron made a transition from the n = 4 to the n = 2 level, what wavelength of light would it emit?



Problem 15: A hydrogen atom is in its second excited state, corresponding to n = 3. Find the radius of the electron’s Bohr orbit

Problem 16: Using the Bohr model, calculate the speed of the electron in a hydrogen atom in the n =

1, 2 and 5 levels.



Problem 17: Consider the Bohr-model description of a hydrogen atom. a) Calculate E2 – E1 and E10 – E9. As n increases, does the energy separation between

adjacent energy levels increase, decrease, or stay the same?

b) Does the radial distance between adjacent orbits increase, decrease, or stay the same as n increase?

c) Show that for any value of n, both Ep = -2Ek and Ek = -E

d) Calculate Ek 1, Ep 1, and E1 for the n = 1 energy level.



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Problem 1: The work function for zinc is 4.31 eV. a) Find the cutoff wavelength for zinc.

b) What is the lowest frequency of light incident on zinc that releases photoelectrons from its surface?

c) If photons of energy 5.50 eV are incident on zinc, what is the maximum kinetic energy of the ejected photoelectrons?

Problem 2: A clean nickel surface is exposed to light of wavelength 206 nm. What is the maximum

speed of the photoelectrons emitted from this surface? The wave functions of nickel is 5.1 eV.

Assignment 2: Photoelectric effect



Problem 3: A beam of light of wavelength 125 nm falls on a metal surface. You observe that the maximum kinetic energy of the ejected electrons is 4.20 eV. Assume that each photon in the beam ejects a photoelectron. What is the work function (in electron volts) of this metal?

Problem 4: The photoelectric threshold wavelength of a tungsten surface is 272 nm. Calculate the

maximum kinetic energy of the electrons ejected from this tungsten surface by ultraviolet

radiation of frequency 1.45 × 1015 Hz. Express the answer in electron volts. Problem 5: When ultraviolet light with a wavelength of 400.0 nm falls on a certain metal surface, the

maximum kinetic energy of the emitted photoelectrons is measured to be 1.10 eV. What is the maximum kinetic energy of the photoelectrons when light of wavelength 300.0 nm falls on the same surface?



Problem 6: The work function for platinum is 6.35 eV. Ultraviolet light of wavelength 150 nm is incident on the clean surface of a platinum sample. We wish to predict the stopping voltage we will need for electrons ejected from the surface. a) What is the photon energy of the ultraviolet light?

b) How do you know that these photons will eject electrons from platinum?

c) What is the maximum kinetic energy of the ejected photoelectrons?

d) What stopping voltage would be required to arrest the current of photoelectrons?



Problem 7: The photoelectric work function of potassium is 2.3 eV. If light having a wavelength of 260 nm falls on potassium, find a) the kinetic energy in electron volts of the most energetic electrons ejected

b) the speed of these electrons.

c) the stopping potential in volts Problem 8: Molybdenum has a work function of 4.20 eV.

a) Find the cutoff wavelength and cutoff frequency for the photoelectric effect.

b) What is the stopping potential if the incident light has a wavelength of 180 nm?



Problem 9: When ultraviolet light with a wavelength of 254 nm falls on a clean copper surface, the stopping potential necessary to stop emission of photoelectrons is 0.181 V. a) What is the photoelectric threshold wavelength for this copper surface?

b) What is the work function for this surface?



Problem 9: Figure below shows the stopping potential versus the incident photon frequency for the photoelectric effect for sodium.

Use the graph to find a) the work function of sodium

b) the planck’s constant

c) the cutoff wavelength



7สียงแล

Problem 1: A photon of green light has a wavelength of 520 nm. Find the photon’s magnitude of

momentum. Additional Problem 1:

Calculate the momentum of a photon whose wavelength is 4.00 × 10-7 m. ___________________________________________________________________________________________

Problem 2: A photon has momentum of magnitude 8.01 × 10-28 kg m/s a) What is the wavelength of this photon? In what region of the electromagnetic

spectrum does it lie?

b) What is the energy of this photon? Give your answer in joules and in electron volts.

Assignment 3: de Broglie's hypothesis



Problem 3: An electron moves with a speed of 4.50 × 106 m/s. What is its de Broglie wavelength? Additional Problem 3:

A proton moves with the speed of 4.50 × 106 m/s. Determine its de Broglie wavelength. ___________________________________________________________________________________________

Problem 4: An electron has a de Broglie wavelength of 2.79 × 10-10 m. Determine a) the magnitude of its momentum

b) its kinetic energy (in joules and in electron volts).



Problem 5: If a photon and an electron each have the same energy of 20.0 eV, find the wavelength of each.

Problem 6: Find the speed of an electron having the same momentum as the photon whose

wavelength is 4.00 × 10-7 m.

Problem 7: An electron is moving with a speed of 8.00 × 106 m/s. What is the speed of a proton that has the same de Broglie wavelength as this electron?



Problem 8: Through what potential difference must electrons be accelerated if they are to have the wavelength of 0.220 nm

Through what potential difference must electrons be accelerated if they are to have the same energy as the x ray of wavelength 0.220 nm

Additional Problem 8: What accelerating potential is needed to produce electrons of wavelength 8.00 nm?



Problem 9: In the Bohr model of the hydrogen atom, what is the de Broglie wavelength of the electron when it is in a) the n = 1 level?

b) the n = 4 level? Additional Problem 9: A hydrogen atom is in its second excited state, corresponding to n = 3. Find the de Broglie wavelength of the electron in this orbit.



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Problem 1: How many protons and how many neutrons are there in a nucleus of the most common isotope of

a) Silicon Si1428

b) Rubidium Rb3785

c) Thallium Tl81205

Problem 2: The most common isotope of uranium, U92238 , has atomic mass 238.050788 u. Calculate

a) the mass defect.

b) the binding energy (in MeV).

c) the binding energy per nucleon. Additional Problem 2.1: Calculate the mass defect, the binding energy (in MeV), and the binding energy per nucleon of

a) the nitrogen nucleus, N714

b) the helium nucleus, He24

c) How does the binding energy per nucleon compare for these two nuclei? Additional Problem 2.2: Calculate a) the total binding energy and

b) the binding energy per nucleon of Cx12 .

Assignment 4: Binding energy



Problem 3: Calculate the binding energy of Cax40 .

Problem 4: The atomic mass of C611 is 11.011434 u, and the atomic mass of B5

11 is 11.009305u.

a) Calculate the binding energies of B511 and C6

11 .

b) Which of the two nuclei has the larger binding energy? Why is this so?



Problem 5: Calculate the binding energy per nucleon for the nuclei Kr3686 and Ta73

180 .

Problem 6: The peak of the graph of nuclear binding energy per nucleon occurs near Fex56 . Show that

Fex56 has a higher binding energy per nucleon than its neighbors Mnx

55 and Cox59 .



7สียงแล

Problem 1: Identify the unknown nuclide or particle (X).

a) X → Ni2865 + γ

b) Po84215 → X + α

c) X → Fe2655 + e+ + ν

Problem 2: What particle (α particle, electron, or positron) is emitted in the following radioactive decays?

a) Si1427 → Al13

27

b) U92238 → Th90

234

c) As3374 → Se34

74

Problem 3: What nuclide is produced in the following radioactive decays?

a) α decay of Pu94239

b) β- decay of Na11

24

c) β+ decay of O8

15

Assignment 5: Radioactive decay



7สียงแล

Problem 1: Tritium has a half-life of 12.33 years. What fraction of the nuclei in a tritium sample will remain a) after 5.00 yr?

b) after 10.0 yr?

c) after 123.3 yr?

Problem 2: The most common isotope of radon is Rnx222 , which has half-life 3.82 days.

a) What fraction of the nuclei that were on the Earth one week ago are now undecayed?

b) Of those that existed one year ago?

Assignment 6: Half-life



Problem 3: A radioactive isotope has a half-life of 43.0 min. At t = 0 its activity is 0.376 Ci. What is its activity at t = 2.00 h?

Problem 4: The radioactive isotope Aux198 has a half-life of 64.8 h. A sample containing this isotope has

an initial activity (t = 0) of 40.0 µCi. Calculate the number of nuclei that decay in the time interval between t1 = 10.0 h and t2 = 12.0 h.



Problem 5: Measurements on a certain isotope tell you that the decay rate decreases from 8265 decays/min to 3088 decays/min in 4.00 days. What is the half-life of this isotope?

Problem 6: In a piece of rock from the Moon, the Rbx87 content is assayed to be 1.82 × 1010 atoms per

gram of material and the Srx87 content is found to be 1.07 × 109 atoms per gram. The

relevant decay relating these nuclides is

Rbx87 → Srx

87 + e- + ν̅

The half-life of the decay is 4.75 × 1010 yr. Calculate the age of the rock.



Problem 7: The unstable isotope Kx40 is used for dating rock samples. Its half-life is 1.28 × 109 y.

a) How many decays occur per second in a sample containing 1.51 × 10-6 g of Kx40 .

b) What is the activity of the sample in curies?

Additional Problem 7:

Consider a 1.00-kg sample of natural uranium composed primarily of Ux238 . Find the activity in curies due to

this isotope. ___________________________________________________________________________________________

Problem 8: The common isotope of uranium, Ux238 , has a half-life of 4.47 × 109 years, decaying to Thx

234 by alpha emission. a) What is the decay constant?

b) What mass of uranium is required for an activity of 1.00 curie?

c) How many alpha particles are emitted per second by 10.0 g of uranium?



Problem 9: The radioactive nuclide Ptx199 has a half-life of 30.8 minutes. A sample is prepared that has

an initial activity of 7.22 × 1011 Bq.

a) How many Ptx199 nuclei are initially present in the sample?

b) How many are present after 30.8 minutes? What is the activity at this time?

c) Repeat part (b) for a time 92.4 minutes after the sample is first prepared. Problem 10: If a 5.63-g sample of an isotope having a mass number of 114 decays at a rate of 0.400 Ci,

what is its half-life?



Problem 11: As a health physicist, you are being consulted about a spill in a radiochemistry lab. The

isotope spilled was 400 µCi of Bax131 , which has a half-life of 12 days.

a) What mass of Bax131 was spilled?

b) Your recommendation is to clear the lab until the radiation level has fallen 1.00 µCi. How long will the lab have to be closed?

Problem 12: A sample of carbon from living matter decays at the rate of 172.5 decays/minute due to

the radioactive Cx14 in it. If the isotope of Cx

14 has a half-life of 5,730 years, what will be the decay rate of this sample in a) 1000 years?

b) 50,000 years?



7สียงแล

Problem 1: Identify the unknown nuclides or particles X in the nuclear reactions.

a) X + He24 → Mg12

24 + n01

b) He92235 + n0

1 → Sr3890 + X + 2( n0

1 )

c) 2( H11 )→ H1

2 + X + X′

Assignment 7: Nuclear reaction



Problem 2: How much energy is released in each reaction?

H11 + B5

11 → 3( He24 )

Problem 3: Calculate the reaction energy for the reaction

p + H13 → H + H1

212

Problem 4: How much energy is released in the reaction given in the following?

N713 → C6

13 + e+ + ν



Problem 5: Consider the nuclear reaction

H12 + Be4

9 → X + He24

where X is a nuclide. a) What are the values of Z and A for the nuclide X?

b) How much energy is liberated? Problem 6: Consider the nuclear reaction

He24 + Li3

7 → X + n01

where X is a nuclide. a) What are Z and A for the nuclide X?

b) Is energy absorbed or liberated? How much?



Problem 7: Consider the nuclear reaction

Si1428 + γ → Mg 12

24 + X where X is a nuclide. a) What are Z and A for the nuclide X?

b) Ignoring the effects of recoil, what minimum energy must the photon have for this

reaction to occur? The mass of a Si1428 atom is 27.976927 u, and the mass of a Mg 12

24 atom is 23.985042 u.



7สียงแล Problem 1: Calculate the energy released in the fusion reaction

He23 + H1

2 → He24 + H1

1 Problem 2: a) Calculate the energy (in electron volts) released in the deuterium-tritium fusion reaction.

H12 + H1

3 → He + n01

24

b) Calculate the energy (in kilowatt-hours) released if 1.00 kg of deuterium undergoes fusions according

to this reaction. Problem 3: Consider the fusion reaction

H12 + H1

2 → He23 + n0

1 a) Compute the energy liberated in this reaction in MeV and in joules.

b) Compute the energy liberated per mole of deuterium.

Assignment 8: Fission & Fusion



Problem 4: The equation of a fission process is given in which Ux235 is struck by a neutron and

undergoes fission to produce Bax144 , Krx

89 and three neutrons. The measured masses of

these isotopes are 235.043930 u ( Ux235 ), 143.922953 u ( Bax

144 ), 88.917631 u ( Krx89 ), and

1.0086649 u (neutron). a) Calculate the energy (in MeV) released by each fission reaction.

b) Calculate the energy released per gram of Ux235 in MeV/g.

Problem 5: Calculate the energy (in kilowatt-hours) released if 1.00 kg of Pux239 undergoes complete

fission and the energy released per fission event is 200 MeV.



Problem 6: The United States uses about 1.4 × 1019 J of electrical energy per year. If all this energy

came from the fission of Ux235 , which releases 200 MeV per fission event,

a) how many kilograms of Ux235 would be used per year?

b) how many kilograms of uranium would have to be mined per year to provide that

much Ux235 ? (Recall that only 0.70% of naturally occurring uranium is Ux

235 )

Problem 7: A nuclear power plant operates by using the energy released in nuclear fission to convert 20 oC water into 400 oC steam. How much water could theoretically be converted to steam

by the complete fissioning of 1.00 g of Ux235 at 200 Mev/fission?

References

1. Young, H. D., Freedman, R. A. (2016). Sears and Zemansky’s University Physics with Modern Physics. (14th ed). Pearson.

2. Serway, R. A., Jewett, Jr., J. W. (2018). Physics for Scientists and Engineers with Modern Physics. (10th ed). Cengage learning.

3. Henderson, H. (2017). SAT Subject Test Physics. (10th ed). Grace Freedson’s Publishing Network.

4. The Staff of the Princeton Review. (2017). Cracking the SAT Subject Test in Physics. (16th ed). TPR Educational IP Holdings.

5. Jansen, R. and Young, G. (2016). BARRON’S SAT Subject Test in Physics. (2nd ed). Barron’s Educational Series.

6. ขวัญ อารยะธนิตกุล, นฤมล เอมะรัตต์, รัชภาคย์ จิตต์อารี และ เชิญโชค ศรขวัญ. (2558). ฟิสิกส์ 2. (ฉบับปรับปรุงครั้งที ่9). ภาควิชาฟิสิกส์ คณะวิทยาศาสตร์ มหาวิทยาลัยมหิดล.

7. สถาบันส่งเสริมการสอนวิทยาศาสตร์และเทคโนโลยี. (2563). หนังสือเรียนรายวิชาเพิ ่มเติมวิทยาศาสตร์ ฟิสิกส์ เล่ม 6. (พิมพ์ครั้งที่ 1). กรุงเทพฯ: โรงพิมพ์ สกสค. ลาดพร้าว.

Biography

Name Date of Birth Place of Birth

Ariyaphol Jiwalak 4 July 1992 Chanthaburi, Thailand

Educational Background Mahidol University, 2011-2014 Bachelor of Science (Physics)

Mahidol University, 2015-2018 Master of Science (Physics, International Program) Royal Government of Thailand Scholarship

Development and Promotion of Science and Technology Talent Project (DPST) by the Institute for the Promotion of Teaching Science and Technology (IPST)

Home Address 5/371 Ideo Mobi Bangsue Grand Interchange Prachachuen Road, Bangsue Sub-district, Bangsue District, Bangkok, 10800

Tel. 089-7525223 E-mail: [email protected] Publication / Presentation Jiwalak A., Emarat N. and Arayathanitkul K., Students' physics laboratory skill in

measurement and uncertainty. Siam Physics Congress, 20-22 May 2015, Krabi, Thailand.

Jiwalak A., Emarat N. and Arayathanitkul K., An activity sheet for teaching double-slit interference. Siam Physics Congress, 21-23 May 2018, Phitsanulok, Thailand.

โรงเรียนสาธิตมหาวิทยาลัยราชภัฏสวนสุนันทาเลขท่ี 1 ถนนอูทองนอก แขวงดุสิต เขตดุสิต

กรุงเทพมหานคร 10300

physics 6 (sci 33202)

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