The Compton Effect is the phenomenon in which the wavelength of an X-ray or gamma-ray photon increases after it is scattered by an electron.
This discovery was made by Arthur H. Compton in 1923 and provided strong evidence that light behaves as a stream of particles called photons.
Experimental setup
The apparatus consisted of:
- A source of monochromatic X-rays.
- A graphite target containing loosely bound electrons.
- A rotating detector (spectrometer) to measure scattered X-rays at different angles \theta.
- A wavelength analyzer to determine the wavelength of the scattered radiation.
Procedure
- A beam of X-rays with wavelength \lambda was directed onto a graphite target.
- The scattered radiation was observed at various angles.
- The wavelength of the scattered X-rays was measured accurately.
Derivation of Compton Shift Formula
Consider a photon of wavelength \( \lambda \) incident on a free electron at rest. After collision, the photon is scattered through an angle \( \theta \) with wavelength \( \lambda' \), and the electron recoils with momentum \( P \).
1. Conservation of Energy
According to the law of conservation of energy, the total energy of an isolated system remains constant. Therefore,
\[ \text{Total Energy Before Collision} = \text{Total Energy After Collision} \]Energy Before Collision
Before the collision, the system consists of:
- An incident photon of wavelength \( \lambda \)
- An electron initially at rest
The energy of the incident photon is
\[ E_{\gamma} = hf = \frac{hc}{\lambda} \]The electron is at rest, so its kinetic energy is zero. However, it still possesses rest-mass energy:
\[ E_e = m_e c^2 \]Hence, the total energy before collision is
\[ E_{\text{before}} = \frac{hc}{\lambda} + m_e c^2 \]Energy After Collision
After the collision:
- The photon is scattered with wavelength \( \lambda' \)
- The electron recoils with momentum \( P \)
The energy of the scattered photon is
\[ E'_{\gamma} = \frac{hc}{\lambda'} \]The recoiling electron now has relativistic energy
\[ E = \sqrt{P^2c^2+m_e^2c^4} \]Therefore, the total energy after collision is
\[ E_{\text{after}} = \frac{hc}{\lambda'} + E \]Applying Conservation of Energy
Since energy is conserved,
\[ E_{\text{before}} = E_{\text{after}} \]Substituting the expressions for the energies:
\[ \boxed{ \frac{hc}{\lambda} + m_e c^2 = \frac{hc}{\lambda'} + E } \]This is the energy conservation equation used in the derivation of the Compton shift formula.
Important Note
Even though the electron is initially at rest, its energy is not zero. Every particle with mass possesses rest energy:
\[ E_0 = m_e c^2 \]Therefore, the electron contributes \(m_e c^2\) to the total energy before the collision.
The total energy before collision equals the total energy after collision:
\[ \frac{hc}{\lambda}+m_e c^2 = \frac{hc}{\lambda'}+E \]where \(E\) is the relativistic energy of the recoiling electron. Rearranging,
\[ E-m_e c^2 = hc\left(\frac{1}{\lambda}-\frac{1}{\lambda'}\right) \tag{1} \]2. Conservation of Momentum
The momentum conservation equation comes from the Law of Conservation of Linear Momentum, which states:
\[ \boxed{ \text{Total momentum before collision} = \text{Total momentum after collision} } \]Since the photon and electron form an isolated system during the collision, their total momentum must remain constant.
Step 1: Momentum Before Collision
Before the collision:
- The photon is moving toward the electron.
- The electron is at rest.
The momentum of the incident photon is
\[ p=\frac{h}{\lambda} \]Since the electron is at rest,
\[ p_e=0 \]Therefore, the total momentum before collision is
\[ \boxed{ \vec{p} } \]Step 2: Momentum After Collision
After the collision:
- The photon is scattered at an angle \( \theta \).
- The electron recoils with momentum \( \vec{P} \).
The scattered photon momentum is
\[ p'=\frac{h}{\lambda'} \]Thus, the total momentum after collision is
\[ \boxed{ \vec{p'}+\vec{P} } \]Step 3: Apply Conservation of Momentum
\[ \boxed{ \vec{p} = \vec{p'}+\vec{P} } \]This is the vector form of the momentum conservation equation used in Compton scattering.
Step 4: Resolve into Components
Take the incident photon direction as the x-axis.
The scattered photon makes an angle \( \theta \) with the x-axis.
The x-component of scattered photon momentum is
\[ p'_x=p'\cos\theta \]The y-component is
\[ p'_y=p'\sin\theta \]x-direction
\[ p = p'\cos\theta + P_x \]Substituting \( p=\frac{h}{\lambda} \) and \( p'=\frac{h}{\lambda'} \):
\[ \boxed{ \frac{h}{\lambda} = \frac{h}{\lambda'}\cos\theta + P_x }\tag{2} \]y-direction
Initially there is no momentum along the y-axis. Therefore,
\[ 0 = p'\sin\theta - P_y \] or \[ \boxed{ P_y = \frac{h}{\lambda'}\sin\theta }\tag{3} \]The magnitude of the electron momentum is related to its components by Pythagoras' theorem:
\[ P^2 = P_x^2+P_y^2 \tag{4}\] Substituting (2) and (3): \[ P^2 = \left( \frac{h}{\lambda} - \frac{h}{\lambda'}\cos\theta \right)^2 + \left( \frac{h}{\lambda'}\sin\theta \right)^2 \] ---Step 3: Expand the Squares
\[ P^2 = \left(\frac{h}{\lambda}\right)^2 + \left(\frac{h}{\lambda'}\cos\theta\right)^2 - 2\frac{h^2}{\lambda\lambda'}\cos\theta + \left(\frac{h}{\lambda'}\sin\theta\right)^2 \] Grouping terms, \[ P^2 = \left(\frac{h}{\lambda}\right)^2 + \frac{h^2}{\lambda'^2} \left(\cos^2\theta+\sin^2\theta\right) - 2\frac{h^2}{\lambda\lambda'}\cos\theta \] Using the trigonometric identity \[ \cos^2\theta+\sin^2\theta=1 \] gives \[ P^2 = \left(\frac{h}{\lambda}\right)^2 + \left(\frac{h}{\lambda'}\right)^2 - 2\frac{h^2}{\lambda\lambda'}\cos\theta \] ---Final Result (Equation 4)
\[ \boxed{ P^2 = \left(\frac{h}{\lambda}\right)^2 + \left(\frac{h}{\lambda'}\right)^2 - 2\frac{h^2}{\lambda\lambda'}\cos\theta } \]This is Equation (4). It represents the square of the recoil electron's momentum in terms of the incident photon momentum, scattered photon momentum, and scattering angle.
Mathematically, it is the same as applying the cosine rule to the momentum vector triangle:
\[ \vec P = \vec p - \vec p' \] which directly gives \[ P^2 = p^2 + p'^2 - 2pp'\cos\theta. \]The experimental success of the Compton effect was one of the strongest proofs that photons really possess momentum.
\[ P^2 = \left(\frac{h}{\lambda}\right)^2 + \left(\frac{h}{\lambda'}\right)^2 - 2\frac{h^2}{\lambda\lambda'}\cos\theta \tag{5} \]3. Relativistic Energy Relation
For the recoiling electron,
\[ E^2 = P^2c^2+m_e^2c^4 \tag{6} \]Substituting Equation (5) into Equation (6):
\[ E^2 = m_e^2c^4 + h^2c^2 \left[ \frac{1}{\lambda^2} + \frac{1}{\lambda'^2} - \frac{2\cos\theta}{\lambda\lambda'} \right] \tag{7} \]4. Eliminating the Electron Energy
From Equation (1),
\[ E = m_e c^2 + hc \left( \frac{1}{\lambda} - \frac{1}{\lambda'} \right) \]Squaring both sides:
\[ E^2 = m_e^2c^4 + 2m_ehc^3 \left( \frac{1}{\lambda} - \frac{1}{\lambda'} \right) + h^2c^2 \left( \frac{1}{\lambda} - \frac{1}{\lambda'} \right)^2 \tag{7} \]Equating Equations (6) and (7) and simplifying:
\[ m_e c \left( \frac{1}{\lambda} - \frac{1}{\lambda'} \right) = \frac{h}{\lambda\lambda'} (1-\cos\theta) \]Multiplying both sides by \( \lambda\lambda' \):
\[ m_e c(\lambda'-\lambda) = h(1-\cos\theta) \]5. Compton Shift Formula
\[ \boxed{ \lambda'-\lambda = \frac{h}{m_e c} (1-\cos\theta) } \]The wavelength shift is therefore:
\[ \boxed{ \Delta\lambda = \lambda'-\lambda = \frac{h}{m_e c} (1-\cos\theta) } \]Compton Wavelength
\[ \lambda_C = \frac{h}{m_e c} = 2.426\times10^{-12}\,\text{m} \]Hence the Compton shift equation can also be written as:
\[ \boxed{ \Delta\lambda = \lambda_C(1-\cos\theta) } \]This is the required expression for the Compton Effect.