Question

Q. A 511 kev gamma ray photon is Compton scattered form a free electorn in an aluminium block. What is wavelength of incident photon? What is wavelength of scattered photon? What is energy of scattered photon?
Q. show that ∆E/E = hf’/moc2(1-cos ө )

Answer 1

Answer on Question #58601-Physics-Quantum Mechanics

Q. A 511 kev gamma ray photon is Compton scattered from a free electron in an aluminum block. What is wavelength of incident photon? What is wavelength of scattered photon? What is energy of scattered photon?

Solution

Energy E of a photon having wavelength $\lambda$ can be written as

$$E = \frac{1240 \, eV \cdot nm}{\lambda}.$$

The wavelength of incident photon is

$$\lambda = \frac{1240 \, eV \cdot nm}{511 \cdot 10^3 \, \mathrm{eV}} = 2.43 \, \mathrm{pm}.$$

The wavelength of scattered photon is

$$\lambda' = \lambda + \frac{h}{m_e c} (1 - \cos \theta), \quad \frac{h}{m_e c} = 2.43 \, \mathrm{pm},$$

where $\theta$ is a scattering angle.

$$\lambda' = 2.43 \, \mathrm{pm} + 2.43 \, \mathrm{pm} (1 - \cos \theta) = 2.43 (2 - \cos \theta) \, \mathrm{pm}$$

The energy of scattered photon is

$$E = \frac{1240 \, eV \cdot nm}{\lambda'} = \frac{1240 \, eV \cdot nm}{2.43 (2 - \cos \theta) \, \mathrm{pm}} = \frac{511}{(2 - \cos \theta)} \, \mathrm{keV}.$$

Q. show that $\Delta E / E = h f' / \mathrm{moc} \cdot 2 \cdot (1 - \cos \theta)$

Solution

$$\lambda' = \lambda + \frac{h}{m_e c} (1 - \cos \theta)$$ $$E' = \frac{h c}{\lambda'}; \quad E = \frac{h c}{\lambda}.$$ $$\frac{\Delta E}{E} = \frac{\frac{h c}{\lambda'} - \frac{h c}{\lambda}}{\frac{h c}{\lambda}} = \frac{\lambda - \lambda'}{\lambda'} = -\frac{\frac{h}{m_e c} (1 - \cos \theta)}{\frac{c}{f'}} = -\frac{h f'}{m_e c^2} (1 - \cos \theta)$$ $$\left| \frac{\Delta E}{E} \right| = \frac{h f'}{m_e c^2} (1 - \cos \theta)$$

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