Question

Q. Photons of wavelength of 2.17pm are incident on free electron (a) find wavelength of photon that is scattered at 350 from the incident direction.

Answer 1

Answer on Question 58655, Physics, Quantum Mechanics

Question:

Photons of wavelength of $2.17~pm$ are incident on free electrons

(a) find wavelength of photon that is scattered at $35{}^{\circ}$ from the incident direction.

(b) do the same if the scattering angle is $115{}^{\circ}$ .

Solution:

In this question, we are dealing with the famous Compton effect. Here's the explanation of the Compton effect:

A photon of wavelength $\lambda$ comes in from left, collides with a free electron at rest, and a new photon of wavelength $\lambda'$ emerges at an angle $\theta$ (in case of (a), $\theta = 35{}^{\circ}$ ; in case of (b), $\theta = 115{}^{\circ}$ ). Part of the energy of the photon is transferred to the recoiling electron (the arrow in the picture indicates the direction of motion of the electron).

a) We can find $\lambda^{\prime}$ from the Compton Scattering equation:

$$\lambda^ {\prime} - \lambda = \frac {h}{m _ {e} c} (1 - \cos \theta),$$

here, $\lambda$ the initial wavelength of the photon, $\lambda^{\prime}$ is the wavelength after scattering, $\frac{h}{m_ec}$ is the Compton wavelength, and it is equal to $2.43\cdot 10^{-12}m$ , $h$ is the Planck's constant, $m_{e}$ is the electron rest mass, $c$ is the speed of light, $\theta$ is the scattering angle.

Therefore, from this formula we can calculate $\lambda'$ for the photon that is scattered at $35{}^{\circ}$ from the incident direction:

$$\begin{array}{l} \lambda' = \lambda + \frac{h}{m_e c} (1 - \cos \theta) = 2.17 \cdot 10^{-12} \, m + 2.43 \cdot 10^{-12} \, m \cdot (1 - \cos 35{}^\circ) = \\ = 2.61 \cdot 10^{-12} \, m = 2.61 \, \text{pm}. \end{array}$$

b) We can calculate $\lambda'$ for the photon that is scattered at $115{}^\circ$ from the incident direction from the same formula:

$$\begin{array}{l} \lambda' = \lambda + \frac{h}{m_e c} (1 - \cos \theta) = 2.17 \cdot 10^{-12} \, m + 2.43 \cdot 10^{-12} \, m \cdot (1 - \cos 115{}^\circ) = \\ = 5.63 \cdot 10^{-12} \, m = 5.63 \, \text{pm}. \end{array}$$

**Answer:**

a) $\lambda'(35{}^\circ) = 2.61 \cdot 10^{-12} \, m = 2.61 \, \text{pm}$.

b) $\lambda'(115{}^\circ) = 5.63 \cdot 10^{-12} \, m = 5.63 \, \text{pm}$.

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