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Chapter 1 Electromagnetic Radiation Behaving as Particles

Particle and wave

Classical Particles

Electromagnetic Waves

Gauss’s law for electricity

Gauss’s law for magnetism

Faraday’s law

Generalized Ampere’s law

E field can be created by changing B field!

B field can be created by changing E field!

Maxwell Equations

Electromagnetic wave propagates at a speed of:

This value coincided with the measured value of the speed of visible light. Visible light was thus attributed to the EM wave.

Hertz’s Experiment Generating EM Waves

Hertz found that waves of electric field and magnetic field were created while there was electric current pulse in a discharge across the gap. The generated wave traveled at the same speed speed as light but a much larger wavelength, and had same properties as light wave, like reflection, refraction, interference and polarization.

Electromagnetic wave spectrum

Principle of superposition: If there are two waves y1(x,t) and y2(x,t) passing through a point x at time t, the resultant wave disturbance at the position x and time t is given by y(x,t)=y1(x,t)+y2(x,t).

Young’s double slit experiment - The observed interference pattern is a characteristic wave phenomenon, which can be deduced from the Superposition Principal. - Classical particle cannot have the similar phenomenon.

Blackbody Radiation

- While charged particles accelerate, EM waves emit. All materials thus emits EM waves because they contain charged particles undergoing random motion. The energy spectrum emitted depends on the average energy of the motion and thus is a function of the temperature.

The body that absorbs all the electromagnetic radiation incident on it is called a blackbody.

Blackbody Radiation

-The intensity dI of the emitted electromagnetic radiation in the frequency range f to f+df:

is called the spectral energy density in volume V.

Measurement of spectrum of thermal radiation

Experimental results of blackbody radiation

The total intensity integrated over all wavelengths was observed to follow:

σ = 5.671×10-8 Wm-2K-4 is called the Stefan Boltzmann constant.

Stefan Boltzmann law

The frequency at which the observed spectral energy density dU/df reaches the maximum follows the Wien’s displacement law :

• Classical law of equipartition of energy:

for system in thermal equilibrium at temperature T.

Note that degree of freedom = number of mode of energy possession. This resulted in the Rayleigh-Jeans formula:

k = 1.381×10-23 JK-1 is call the Boltzmann constant.

Failure of Classical wave theory at high frequency: the ultraviolet catastrophe

Planck’s suggestion: An oscillating atom can only absorb or emit energy only in

discrete bundles called quanta:

is called the Plank constant.

The so derived Plank’s formula fitted well with the experimental observation.

The Photoelectric effect

When a metal surface is illuminated by light, electron may be emitted from the surface. This phenomenon is known as the photoelectric effect.

Stopping potential VS is the minimum ΔV needed to stop the photocurrent.

The maximum KE of the emitted electron is the given by: KEmax = eVS

Work function of the metal Φ = the mimimum energy required to remove an electron from the metal

(1) For a fixed wavelength of light illumination, the maximum KE of electron emitted as derived from measuring the stopping potential is independent of the light source intensity. Doubling the light intensity for the same wavelength exactly double the current.

(2) There is no electron emitted at all if the light frequency of the light is below a value called the cut off frequency fc. This cut-off frequency is independent of the light intensity. This violates the classical wave theory.

Einstein idea of photon Absorption of light energy in form of discrete energy bundle called photon with the energy of:

E=hf

- For a fixed wavelength f, each of the photon carry the same energy hf. Increasing the light intensity while fixing the wavelength implies increasing the number of photons having the same energy hf. - For photons hf incident on the metal: KE<hf-Φ, or there exists a max. KE such that KEmax=hf-Φ. - By putting KEmax=0, we can obtain the cut off frequency:

Production of X-ray X-ray are produced while the electrons are accelerated through high voltage ~10keV, and then bring to stop by striking on a metal target like Cu.

It was observed the existence of a minimum wavelength λc such that no X-ray having wavelength smaller than λc would emit. Could not be explained by classical wave theory.

The X-ray photon emitted during the collision: hν=K-K’

The emission of X-ray photon having the minimum wavelength λc (i.e. highest frequency) corresponds to the complete conversion of the electron’s KE to the X-ray photon, i.e.

The electron striking into the metal are brought to stop by numerous collisions with the atoms.

Compton Effect

According to classical wave theory, Light wave entering a matter would accelerate the electrons inside the material, and thus electromagnetic wave having the same frequency would re-radiate. The scattering process thus results in decreasing the intensity while having the wavelength unchanged.

However, Compton observed an angle dependent decrease of the scattered X-ray wavelength.

According to the special relativity, an object having a zero mass has the momentum of:

Integrating with the photon theory:

Interaction of a X-ray photon with an electron initially at rest

Einitial = Efinal

(1)

(2)

(3)

(2) and (3):

The relativistic relation between the electron’s energy and momentum:

(4)

(5)

Sub. (1) and (4) in (5):

(1) Notice that λ’-λ≥0, implying the scattered photon has a larger wavelength, or is red shifted (i.e. lower photon energy).

(2) X-ray particle behaves like particle having momentum p=hf/c.

Relativity:-

Relativistic energy and momentum:

For a stationary mass:

€

KE = (Energy moving) - (Energy at rest)= γ umc

2 −mc 2

= 1- u2

c 2

−1/ 2

−mc 2

If u << c, KE ≈12mv 2

If u<<c or KE<<mc2, then we do not need to use the relativistic approach.

Pair Production

A photon having the energy of hν converted to an electron-positron pair. Positron is the anti-particle of electron, having the identical mass but opposite charge to electron.

Conservation of relativistic energy:

Since K+ and K- are positive, hν ≥ 2 me c2 =1.02MeV the photon giving rise to a pair production process is in the range of gamma ray.

Remark: Pair production cannot occur without the presence of an atom, which would carry the recoil momentum for fulfilling the momentum conservation.

Annihilation

Annihilation process is the conversion of a particle and its anti-particle to photon. An example is:

e- + e+ 2γ’s photons

There must be more than one gamma photon being produced in order to fulfill the momentum conservation.

Conceptual question 1

What assumptions did Planck make in dealing with the problem of blackbody radiation? Discuss the consequences of the assumptions.

Conceptual question 1 (cont.)

Answer: Planck made two assumptions: (1) Energy of radiating oscillator is quantized and (2) they emit or absorb energy in discrete irreducible packets. Here the ''oscillator'' refers to the atoms or molecules that made up the wall of the blackbody cavity. These assumptions contradict the classical ideas that energy is continuously divisible.


Consider the following properties of the photoelectric effect: (i) The generation of photoelectrons. (ii) The existence of a threshold frequency. (iii) The photoelectric current increases with increasing light intensity. (iv) The photoelectric current is independent of anode- cathode potential difference ΔV for ΔV > 0. (v) The photoelectric current decreases slowly as ΔV becomes more negative. (vi) The stopping potential is independent of the light intensity.


(vii) The photoelectric current appears instantly when the light is turned on. Which of these cannot be explained by classical physics?


Answer: (ii), (vi) and (vii) cannot be explained by classical physics.


The figure shows a typical current-versus-potential difference graph for a photoelectric effect experiment. On the figure, draw and label curves for the following three situations: (i) The light intensity is increased. (ii) The light frequency is increased (but no change in density of photon). (iii) The cathode work function is increased. In each case, no other parameters of the experiment are changed.


Answer:

(i) Maximum current ↑

(ii) |Vstop| ↑ (iii) |Vstop| ↓


Must Compton scattering take place only between X-rays and free electrons? Can radiation in the visible region (say, a green light) undergo Compton scattering with a free electron?


Answer: The wavelength of visible light (e.g. λgreen ≈ 10-6 m) is much larger than the Compton wavelength of the electron (λe = h/mec ≈ 10-12 m). For visible light, the Compton shift is negligibly small compared with its wavelength. So the change in wavelength of the visible light due to the Compton scattering would be too small to be measured.


A beam of photons passes through a block of matter. What are the three ways discussed in this chapter that the photons can lose energy in interacting with the material?


Answer: Photoelectric effect, Compton scattering and pair production.

Conceptual question 6 You have a monoenergetic source of X-rays of energy 84 keV, but for an experiment you need 70 keV X-rays. How would you convert the X-ray energy from 84 to 70 keV?


Answer: To convert the X-ray energy from 84 to 70 keV, the X-ray photons need to lose 14 keV energy. The Compton scattering is the chief means by which X-rays lose energy when they pass through matter. Moreover, a scattered X-ray photon typically lose energy in the range of few eVs to 1MeV. So we can convert the X-ray energy by passing the X-ray beam on a block of matter. Then we adjust the scattering angle until the energy of the scattered X-ray photons have lost the desired amount of energy.


The intensity of a beam of light is increased but the frequency of the light is unchanged. As a result: (i) The photons travel faster. (ii) Each photon has more energy. (iii) The photons are larger. (iv) There are more photons per second. Which of these (perhaps more than one) are true? Explain.


Answer: Only (iv) is true. A more intense light delivers more light quanta to the surface and so ejects more photoelectrons per second in the photoelectric effect. Therefore, light with a greater intensity means more photons per second.