Problem

A $50 \, \mathrm{keV}$ X-ray is scattered through an angle of $90^\circ$. Calculate the energy of the X-ray after Compton scattering.

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Data

• Scattering angle: $\theta = 90^\circ$
• Rest mass of electron: $m_0 = 9.11 \times 10^{-31} \, \mathrm{kg}$
• Speed of light: $c = 3 \times 10^8 \, \mathrm{ms}^{-1}$
• Initial photon energy: $E = hf = 50 \, \mathrm{keV}$
• Rest mass energy of an electron:

$$m_0c^2 = 511 \, \mathrm{keV}$$

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Prerequisite Concepts

The energy of a photon after Compton scattering is calculated using:

$$\frac{1}{hf'} = \frac{1}{hf} + \frac{1}{m_0c^2}(1 - \cos \theta)$$

where:

• $hf'$: Energy of the scattered photon.
• $hf$: Energy of the incident photon.
• $m_0c^2$: Rest mass energy of the electron.
• $\theta$: Scattering angle.

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Solution

Step 1: Rest Mass Energy of Electron

The rest mass energy of the electron is:

$$m_0c^2 = 511 \, \mathrm{keV}$$

Step 2: Substitute into the Compton Equation

Using the formula:

$$\frac{1}{hf'} = \frac{1}{hf} + \frac{1}{m_0c^2}(1 - \cos \theta)$$

Substitute values:

$$\frac{1}{hf'} = \frac{1}{50 \, \mathrm{keV}} + \frac{1}{511 \, \mathrm{keV}}(1 - \cos 90^\circ)$$

Since $\cos 90^\circ = 0$:

$$\frac{1}{hf'} = \frac{1}{50} + \frac{1}{511}$$

Step 3: Simplify

Calculate each term:

$$\frac{1}{50} = 0.02 \, \mathrm{keV}^{-1}, \quad \frac{1}{511} \approx 0.00196 \, \mathrm{keV}^{-1}$$

Add the terms:

$$\frac{1}{hf'} = 0.02 + 0.00196 = 0.02196 \, \mathrm{keV}^{-1}$$

Take the reciprocal:

$$hf' = \frac{1}{0.02196} \approx 45.5 \, \mathrm{keV}$$

Step 4: Calculate Energy Transferred to the Electron

The kinetic energy transferred to the electron is the difference between the initial and scattered photon energies:

$$\mathrm{KE}_{\text{electron}} = 50 \, \mathrm{keV} - 45.5 \, \mathrm{keV} = 4.5 \, \mathrm{keV}$$

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Answer

• Energy of the scattered X-ray: $45.5 \, \mathrm{keV}$
• Kinetic energy of the electron: $4.5 \, \mathrm{keV}$

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