Waves can exhibit particle-like characteristics, and particles can exhibit wave-like characteristics. This is called wave-particle duality.

All objects radiate electromagnetic waves. Very hot objects emit e-m waves in the visible spectrum. Cooler bodies emit invisible e-m waves in the infrared region of the spectrum.

A blackbody at a constant temperature absorbs, then reemits all the e-m radiation falling on it. The hotter an object is the more its Planck curve peak shifts toward the visible part of the spectrum.

Planck calculated the blackbody radiation curves using atomic oscillators. He assumed that the energy E of an atomic oscillator could have only the discrete values of E = 0, hf, 2hf, 3hf, etc.In other words, E = nhf. h is Planck’s constant:h = 6.626 0755 x 10-34 J•s.

The energy E is said to be quantized. E-M energy occurs as a collection of discrete amounts or packets of energy, the energy of a packet is equal to hf.

Thus, hf is the smallest packet of energy, and it can be increased only in multiples of hf. Einstein specifically stated that light consists of such packets of energy.

Electrons are emitted from a metal surface when light is shined on it. This is the photoelectric effect. These electrons are called photoelectrons.

The electrons are directed toward a positive electrode and cause a current flow. Einstein explained the photoelectric effect in 1905.

He proposed that light of frequency f consisted of discrete energy packets (photons) with energy equal to Planck’s constant times the frequency, E = hf.

Ex 1 - A sixty-watt incandescent light bulb operates at about 2.1% efficiency. If all the light is green light (l = 555 nm), determine the number of photons per second given off by the bulb.

A photon can give up its energy to an electron in the metal. If the photon has enough energy to do the work of removing the electron, the electron is emitted. The minimum energy needed is the work function W0.

If the photon has more energy than needed to emit the electron, the excess appears as kinetic energy of the electron. The least strongly held electrons end up with maximum kinetic energy KEmax.

Einstein expressed the relationship in this equation: hf = KEmax + W0, or KEmax = hf - W0. At the cutoff frequency f0, the electrons have no KE. KEmax = 0, so W0 = hf.

Below this frequency f0, no electrons will be emitted, even at extreme intensities of light.

Ex 2 - The work function for a silver surface is W0 = 4.73 eV. Find the minimum frequency that light must have to eject electrons from this surface.

The maximum kinetic energy of the photoelectrons at a certain frequency is always the same, even if the light intensity increases.

KEmax = hf - W0

Increasing the intensity causes more electrons to be emitted, but their maximum energy does not increase.

All these facts are more consistent with a particle nature than a wave nature.

A photon is not a normal particle, a normal particle has mass and can’t travel at the speed of light.

______E = mc2/√(1- v2/c2) is the formula for total energy of a particle. This can be rewritten as:

______E•√(1- v2/c2) = mc2

_______√ (1- v2/c2) = zero, because v = c.

_______E•√ (1- v2/c2) = zero, so mc2 , and therefore m = zero, and the photon has no mass.

Arthur Compton found that when an x-ray (energy hf) collides with a stationary electron, the electron recoils with an energy KE and the photon is scattered with a lower energy hf’ (and a lower frequency).

The amount of energy lost by the photon is equal to that gained by the electron. Also, the momentum of the incident photon = the momentum of the scattered photon + the momentum of the recoil electron. (The momentum is related to the angle θ at which the photon scatters.)

The momentum of a photon is given by p = E/c. But E = hf and λ = c/f. Therefore, the momentum of a photon: p = hf/c = h/ λ.From this: λ’ - λ = (h/mc)(1 - cosθ)

λ’ - λ = (h/mc)(1 - cosθ) λ’ is the wavelength of the scattered photon,λ is the wavelength of the incident photon,θ is the scattering angle, m is the mass of the electronh/mc is called the Compton wavelength of the electron, h/mc = 2.43 x 10-12 m.

λ’ - λ = (h/mc)(1 - cosθ)

Cosθ varies from -1 to 1, so λ’ - λ can vary from zero to 2h/mc.

The Compton effect, with its conservation of momentum, provides evidence (along with the photoelectric effect) for the particle nature of light. Light is now accepted to have a dual nature.

Ex 3 - Should a solar sail have a dark surface or a light surface?

Louis de Broglie suggested that if waves can behave as particles, then particles can have a wave nature.

The wavelength λ of a particle is given by λ = h/p.

λ = h/ph is Planck’s constant,p is the relativistic momentum of the particle,and λ is the de Broglie wavelength of the particle.

Experiments have confirmed the wave nature of moving particles (including a Young’s double slit experiment diffracting electrons); however, the effects are observable only for very small particles (subatomic).

Ex 4 - Determine the de Broglie wavelength for (a) an electron (m = 9.1 x 10-31 kg) moving at a speed of 6.0 x 106 m/s and (b) a baseball (mass = 0.15 kg) moving at a speed of 13 m/s.

Particle waves are waves of probability. Bright fringes indicate areas where particles have a high probability of striking, low probability for dark fringes.

Using these concepts, Erwin Schrödinger and Werner Heisenberg developed the new branch of physics called quantum mechanics.

Where particles strike a screen is no longer definite, as with Newtonian physics, but is only a matter of probability. We can predict where most particles will go, but not where a single particle will go.

The y-component of the momentum py of a particle is uncertain. It can vary within the range that places particles within the bright fringe (from zero to ∆pv).

It is possible to relate ∆pv to the width of the slit: sin θ = λ/W tells the angle that specifies the location of the first dark fringe.

If θ is small, sin θ ≈ ∆py/px= λ/W, but px = h/ λ.So, ∆py/px= ∆py/h/ λ ≈ λ /W.Therefore, ∆py ≈ h/W.

∆py ≈ h/W The smaller the slit width, the larger the uncertainty in the y component of the momentum.

The uncertainty of ∆py is related to the uncertainty of the position of the particle as it moves through a slit. The uncertainty of this y position ∆y = W.

So, ∆py ≈ h/W, becomes ∆py ≈ h/∆y, or ∆py•∆y ≈ h. This inability to know the particle’s momentum if you know the initial location of the particle is the Heisenberg uncertainty principle.

Momentum and position: ∆py•∆y > h/2π∆y = uncertainty in a particles y position, ∆py = uncertainty in the y component of the particle’s linear momentum. As one of these increases, the other must decrease.

Heisenberg Uncertainty Principle for Energy and time:∆E•∆t > h/2π

Ex 5 - Assume the position of an object is known so precisely that the uncertainty in the position is only ∆y = 1.5 x 10-11 m. (a) Determine the minimum uncertainty in the momentum of the object. Find the corresponding minimum uncertainty in the speed of the object, if the object is (b) an electron (mass = 9.1 x 10-31 kg) and (c) a Ping-Pong ball (mass = 2.2 x 10-3 kg).