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CHAPTER 2 ATOMIC STRUCTURE2.1 Bohrs Atomic Model 2.2 Quantum Mechanical Model 2.3 Electronic Configuration

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CHAPTER 2 ATOMIC STRUCTURE2.1 Bohrs Atomic Model

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2.1

Bohrs Atomic Model

Learning Outcomes a) Describe Bohrs atomic model. b)Explain the existence of energy levels in an atom. c) Calculate the energy of an electron using: En = -RH(1/n2) , RH = 2.18 x 10-18 J of line spectrum of

d)Describe the formation hydrogen atom.

e) Calculate the energy change of an electron during transition using: E = RH(1/n12 _ 1/n2 2) where RH = 2.18 x 10-18 J4 f) Calculate the photon of energy emitted by an

2.1

Bohrs Atomic Model

Learning Outcomes g)Perform calculation involving the Rydberg equation for Lyman, Balmer, Paschen, Brackett and Pfund series.

1/ = RH(1/n12 _ 1/n2 2) where RH = 1.097 x 107 m-1 and n1 the frequency of the first line of the Balmer series.

Line spectrum Balmer Series Lyman Series

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2.1Example

Bohrs Atomic Model

Which of the line in the Paschen series corresponds to the longest wavelength of photon? Describe the transition that gives rise to the line. Solution : Line A. The electron moves from n=4 to n=3.37

Next Lecture

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2.1 Bohrs Atomic Model The Rydberg Equation

The wavelength emitted by the transition of electron between two energy levels is calculated by using the Rydberg Equation.

1 1 = 1 R H 2 2 n2 n1Where RH = 1.097 107 m-1 = wavelength Since should have a positive value thus n1 < n239

2.1 Bohrs Atomic Model The Rydberg EquationDetermine a) The wavelength in nm b) The frequency c) The energy that is associated with the second line in the Balmer series of the hydrogen spectrum.

Solution (a) Second line of Balmer series: the transition of electron is from n = 4 to n = 2 So, n2=4 and

1 1

=

1

1

2

1 22 ) 1

= (1.097 107

1 22

1 42

= 4.86 107= 486

109 nm 1 m40

2.1 Bohrs Atomic Model The Rydberg EquationExample :

Calculate the wavelengths of the fourth line in the Balmer series of hydrogen.

n1= 2 n2 = 6 RH = 1.097 x 107m-1

1

= RH

1

1

22

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= 4.10 x 10-7 m41

2.1 Bohrs Atomic Model The Rydberg EquationDifferent values of RH and its usage 1. RH = 1.097 107 m-1 2. 3. 1 = RH 4. 5. 6. 7. 8. 9. RH = 2.18 x 10-18 J 10. E = R

12 n1

12 n2

n1 < n2

1 1 2 H n2 n f i

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Sampai sini

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EXAMPLE 2

Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom.

1

1 1

= RH

12 n11

12 n21 52

hc E = E = (6.63 10-34 Js)X(3.00108 ms-1 ) X (0.2303 X 107 m-1 )

=1.097 x 107

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E = 4.58 x 10-19 J = 0.2303 X 107 m-1(or by directly using the energy difference formula)44

Exercise 6Calculate what is; i ) Wavelength ii ) Frequency iii ) Wave number of the last line of hydrogen spectrum in Lyman series (Wave number = 1/wavelength)

For Lyman series; n1 = 1 & n2 = Ans: i.9.116 x10-8 m ii.3.29 x1015 s-1 iii.1.0970 X 107 m-145

SOLUTION1 =1.097 x 107 1 1

i)

12

2

= 1.097 X 107 m-1

= 9.116 X 10-8 m

ii) V = c =3.00 X 108 ms-1 = 3.29 X 1015 s-1 9.116 X 10-8 miii) Wave number =1

= 1.097 X 107 m-146

Ionization Energy Definition : Ionization energy is the minimum energy required to remove one mole of electron from one mole of gaseous atom/ion. M (g) M+(g) + e H = +ve

The hydrogen atom is said to be ionised when electron is removed from its ground state (n = 1) to n = .

At n = , the potential energy of electron is zero, here the nucleus attractive force has no effect on the electron (electron is free from nucleus). 47

Examplen1 = 1, n2 =

E

= RH (1/ni2 1/nf2) = 2.18 X 10 -18 (1/12 1/ 2) = 2.18 X 10 -18 (1 0) = 2.18 X 10 -18 J

Ionisation energy = 2.18 X 10 -18 J x 6.02 X 1023 mol-1 = 1.312 x 106 J mol-1 = 1312 kJ mol-1

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Finding ionisation energy experimentally:Convergent limit 1 st line

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Ionisation energy is determined by detecting the wavelength of the convergence point49

Example

10.97 10.66

10.52

10.27

9.74

8.22

wave number (x106 m-1 )

The Lyman series of the spectrum of hydrogen is shown above. Calculate the ionization energy of hydrogen from the spectrum.

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Solution = hc/ =h x c / = h x c x wave no. = 6.626 x 10-34 J s x 3 x 108 m s-1 x 10.97x 106 m-1 = 218.06x 10-20 J = 2.18 x 10-18 J

E

Ionisation energy = 2.18 X 10 -18 x 6.02 X 1023 J mol-1 =1.312 x 106 J mol-1 = 1312 kJ mol-1

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The weakness of Bohrs Theory Bohr was successful in introducing the idea of quantum energy and in explaining the lines of hydrogen spectrum. His theory could not be extended to predict the energy levels and spectra of atoms and ions with more than one electron. His theory can only explain the hydrogen spectrum or ions contain one electron eg: He+, Li2+. Modern quantum mechanics retain Bohrs concept of discrete energy states and energy involved during transition of electrons but totally reject the circular orbits he introduced.52

Point to PonderDavisson & Germer observed the diffraction of electrons when a beam of electrons was directed at a nickel crystal. Diffraction patterns produced by scattering electrons from crystals are very similar to those produced by scattering X-rays from crystals. This experiment demonstrated that electrons do indeed possess wavelike properties.

Thus, can the specified???

position

of

a

wave

be

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de Broglies PostulateIn 1924 Louis de Broglie proposed that not only light but all matter has a dual nature and possesses both wave and corpuscular properties. De Broglie deduced that the particle and wave properties are related by the expression:

=

h = Planck constant (J s) m = particle mass (kg) = velocity (m/s) = wavelength of a matter wave54

Heisenbergs Uncertainty PrincipleIt is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain.

Stated mathematically,

x p

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Where x = uncertainty in measuring the position p = uncertainty in measuring the momentum = mv h = Planck constant55