Like all EM waves, light transports energy across space. The intensity
(energy per unit area and unit time) is proportional to the square of the
amplitude of the electric field of the light wave. This energy, however,
arrives at a receiver not continuously but in discrete units called
**photons**. The energy transported by an
electromagnetic wave is not continuously distributed over the wave front. It is
transported in discrete packages. **Photons are the
particles of light.**

- Photons always move with the speed of light.
- Photons are electrically neutral.
- Photons have no mass, but they have energy E =
hf = hc/λ.
Here h = 6.626*10
^{-34 }Js is a universal constant called**Planck's constant**. The energy of each photon is inversely proportional to the wavelength of the associated EM wave. The shorter the wavelength, the more energetic is the photon, the longer the wavelength, the less energetic is the photon.- A laser beam and a microwave beam can carry the same amount of energy. In this case the laser beam contains a smaller number of photons, but each photon in the laser beam has a higher energy than the photons in the microwave beam.

- Photons can be created and destroyed. When a source emits EM waves, photons are created. When photons encounter matter, they may be absorbed and transfer their energy to the atoms and molecules. Creation and destruction of photons must conserve energy and momentum. The magnitude of the momentum of a photon is p = hf/c = h/λ.

**Problem:**

- What is the energy of a photon of blue light (λ
= 450 nm) and of a photon of red
light (λ =
700 nm) in units of eV =
1.6*10
^{-19 }J?Solution:

E = hc/λ.

Blue light: E = (6.626*10^{-34 }Js)(3*10^{8 }m/s)/(450*10^{-9 }m) = 4.4*10^{-19 }J = 2.76 eV

Red light: E = (6.626*10^{-34 }Js)(3*10^{8 }m/s)/(700*10^{-9 }m) = 2.8*10^{-19 }J = 1.8 eV

**Problem:**

- A 100-W incandescent light bulb uses 100 W (100 W = 100 J/s) of electrical
power but only radiates about 15 W of actual visible light. Roughly
how many visible photons per second hit the open pages of a typical
hardcover book if the pages are about 2 m from the bulb and face it
directly?
- Solution:

We have to make a variety of estimations and approximations to solve this problem.

To find the number of photons hitting the pages each second, we have to know the light energy hitting the pages per second and the energy per photon. We could compute the latter if we knew the wavelength of the light, but the visible light emitted by a normal incandescent bulb is a mix of wavelengths.

Let us estimate that the average wavelength of the visible light is about 550 nm, which is in the yellow region of the spectrum. This means that the average energy per photon is aboutE = hc/λ = (6.626*10

^{-34 }Js)(3*10^{8 }m/s)/(550*10^{-9 }m) = 3.61*10^{-19 }J = 2.26 eV.To find the number of photons hitting the pages of a book, we need to know the energy per second that falls on the pages. Let us assume that the light energy from the bulb travels uniformly in all directions. Imagine a sphere centered on the light bulb with a radius R = 2 m. Every second, 15 J or so of visible light energy crosses this sphere. If the light energy is spread uniformly over the sphere, then the intensity of the visible light or the energy per unit area per unit time in units of W/m

^{2}at any point on the inner surface of the sphere will beI = P

_{lamp}/(4πR^{2}),where P

_{lamp}is the power of the visible light emitted by the lamp (~15 W) and 4πR^{2}is the area of the inner surface of the sphere.I = (15/(16π))(W/m

^{2})The book pages essentially lie on the inner surface of this imaginary sphere, they have area A. The energy per unit time that falls on on this area is

P

_{book}= IA.

(Energy per unit time falling on area A = energy per unit time per unit area times area A.)Taking a typical hardcover book and measuring it with a ruler, we find that the two facing pages form a rectangle about 18 cm tall and 24 cm wide, so A = (0.18 m)(0.24 m) = 0.058 m

^{2}. SoP

_{book}= (15/(16π))*0.058 W = 0.017 J/s.Finally, to find the number of photons per second falling on these pages, we have to divide P

_{book}by the energy per photon.Number of visible photons from bulb hitting the book per second

= (0.017 J/s)/(3.61*10

^{-19 }J) = 4.7*10^{16}/s.This is not an accurate number, but a useful, order of magnitude estimate.

- Solution:

In 1905, **Einstein**
used the discrete nature of light to explain the **
photoelectric effect**. To demonstrate this effect light is shone on a
metal surface. If the frequency of the light is higher than the
**cutoff frequency** f_{c}, then
electrons are released. No photoelectric electrons are emitted if the frequency
of the light falls below this cutoff frequency f_{c}. For many metal
surfaces the frequency of blue light is greater than f_{c} and the
frequency of red light is less than f_{c}. If red light is shone on the
surface, no electrons are emitted, no matter what the intensity of the light.
If blue light is shone on the surface, electrons are emitted. The number of
emitted electrons depends on the intensity of the light. But even if the
intensity is reduced to a very low value, electrons are still emitted, albeit at
a very low rate.

The photoelectric effect cannot be understood within the wave picture of
light. To eject an electron from a metal surface a certain amount of energy
Φ, called the **work function of the metal**, must be
supplied to this electron. In the wave picture the energy of the light beam
does not depend on the frequency, but only on the intensity, which is
proportional to the square of the amplitude. Einstein explained
the photoelectric effect by postulating that an electron can only receive the
large amount of energy necessary to escape the metal from the EM wave by
absorbing a single photon. If this photon has enough energy, the electron is
freed. Excess energy appears as kinetic energy of the electron. The
maximum kinetic
energy of the electron is given by E =
hf - Φ. If the photon does not have enough
energy, then the electron cannot escape the metal.

Link:

**Problems:**

- Molybdenum has a work function of 4.2 eV. Find the cutoff frequency and
cutoff wavelength for the photoelectric effect.
- Solution:

hf_{c }= Φ,

f_{c }= Φ/h = (4.2 eV)(1.6*10^{-19 }J/eV)/(6.626*10^{-34 }Js) = 1.01*10^{15 }Hz is the cutoff frequency

λ_{c }= c/f_{c }= 296 nm is the cutoff wavelength.

- Solution:
- Electrons are ejected from a metal surface with speeds ranging up to 4.6*10
^{5 }m/s when light with a wavelength of 625 nm is used.(a) What is the work function of the surface?

(b) What is the cutoff frequency of the surface?- Solution:
- (a) Φ =
hf - E. E
= (1/2)mv
^{2 }= (1/2)(9.1*10^{-31 }kg)(4.6*10^{5 }m/s)^{2 }= 9.63*10^{-20 }J

hf = hc/λ = (6.626*10^{-34}Js)(3*10^{8 }m/s)/(625*10^{-9 }m) = 3.18*10^{-19 }J

Φ = (3.18*10^{-19 }J) - (3.18*10^{-19 }J) = 2.21*10^{-19 }J is the workfunction of the surface. - (b) f
_{c }= Φ/h = (2.21*10^{-19 }J)/(6.626*10^{-34 }Js) = 3.35*10^{14 }Hz is the cutoff frequency.

- (a) Φ =
hf - E. E
= (1/2)mv

- Solution:
- Light of wavelength 500 nm is incident on sodium with a work function of
2.28 eV. What is the maximum kinetic energy of the ejected
photoelectron?
- Solution:

E = hf - Φ = hc/λ - Φ.

hf = hc/λ = (6.626*10^{-34 }Js)(3*10^{8 }m/s)/(500*10^{-9 }m) = (3.98*10^{-19 }J)(1 eV)/(1.6*10^{-19}J) = 2.48 eV.

E = 0.2 eV.

- Solution:

Link:

The **Compton effect** or **
Compton scattering** (discovered in 1923, Nobel
Prize in Physics in 1927) is another famous experiment
supporting the existence of photons. To observe the Compton effect, x-rays
are scattered from (nearly) free electrons. To understand Compton
scattering we must think of photons as particles.

Imagine an incoming photon scattering off a stationary electron.

After the collision the photon's and the electron's trajectory will be at some angles with respect to the direction of the incoming photon. If we know the energy and momentum of the incoming photon, and the angle the outgoing photon's trajectory makes with the direction of the incoming photon, then we can determine the outgoing photon's energy using energy and momentum conservation.

The energy of a photon is E = hf = hc/λ and the
momentum of the photon is p = hf/c = h/λ = E/c.

The relativistically correct expression for the energy of the electron is E_{e}
= (p_{e}^{2}c^{2} + m_{e}^{2}c^{4})^{1/2}.

The initial momentum of the electron is (nearly) zero.

Energy conservation requires E_{i} + E_{ei} = E_{f} + E_{ef},
or E_{i} + m_{e}c^{2} = E_{f} + (p_{ef}^{2}c^{2}
+ m_{e}^{2}c^{4})^{1/2}.

Momentum conservation requires **p**_{i }= **p**_{f} + **
p**_{ef}. This is a vector equation.

Rewriting these equations we have

(i) (p_{ef}^{2}c^{2} + m_{e}^{2}c^{4})^{1/2}
= E_{i} + m_{e}c^{2} - E_{f},

(ii) p_{ef}^{2} = **p**_{ef}•**p**_{ef}
= (**p**_{i }- **p**_{f})•(**p**_{i
}- **p**_{f}) = p_{i}^{2} + p_{f}^{2}
- 2**p**_{i}•**p**_{f} = p_{i}^{2}
+ p_{f}^{2} - 2p_{i}p_{f}cosθ,

where θ is the photon scattering angle.

Substituting (ii) into (i) and squaring we find

p_{i}^{2} + p_{f}^{2} - 2p_{i}p_{f}cosθ
+ m_{e}^{2}c^{2} = (E_{i}/c + m_{e}c - E_{f}/c)^{2}
= (p_{i} + m_{e}c - p_{f})^{2}_{
}=_{ }p_{i}^{2} + p_{f}^{2} - 2p_{i}p_{f}
+ m_{e}^{2}c^{2} + 2(p_{i} - p_{f})m_{e}c,

or

p_{i}p_{f}(1 - cosθ ) = (p_{i}
- p_{f})m_{e}c,

(1 - cosθ)/(m_{e}c) = 1/p_{f} - 1/p_{i}.

It is traditional to rewrite this expression in terms of the change of the
photon's wavelength.

Δλ = (λ_{f} -
λ_{i}) = [h/(m_{e}c)](1 - cosθ).

The quantity h/(m_{e}c) = 0.00243 nm = 2.43 pm is called the Compton wavelength
for photon-electron collisions.
The Compton wavelength is the amount by which the photon's wavelength changes
when it scatters through 90°. The change in wavelength is smaller if the
scattering angle is less than 90°, and is a maximum of twice the Compton
wavelength if the photon scatters backwards through 180°. The Compton
wavelength for photon-electron collisions is small (~10^{-12} m) and the Compton effect is only observable with very short
wavelength photons. Only then does the change in the wavelength
Δλ become an appreciable fraction of the wavelength
λ.

**Problem:**

Photons of wavelength λ scatter elastically on
free protons initially at rest. By how much does the wavelength of the
photons scattered at 90^{o} increase?

- Solution:

Δλ = (λ_{f}- λ_{i}) = [h/(m_{p}c)](1 - cosθ), where m_{p}is the mass of the proton.

For θ =90^{o}, Δλ = h/(m_{p}c) = 1.32 fm

For photon-proton collisions the Compton wavelength is even smaller small (~10^{-15} m) and the Compton effect is harder to measure and verify
experimentally.