In the Compton scattering process,the inciden | vedclass

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(D) For the Compton effect,the shift in wavelength is given by the relation:$\Delta \lambda = \lambda_2 - \lambda_1 = \frac{h}{m_0 c}(1 - \cos 	heta)

Vedclass · Accepted Answer

$0.208$

Vedclass · Answer

(D) For the Compton effect,the shift in wavelength is given by the relation:$\Delta \lambda = \lambda_2 - \lambda_1 = \frac{h}{m_0 c}(1 - \cos 	heta)$Here,$\lambda_1$ is the wavelength of incident radiation,$\lambda_2$ is the wavelength of scattered radiation,and $\frac{h}{m_0 c}$ is the Compton wavelength,which is approximately $0.024 \ \mathring{A}$.Given: $\lambda_2 = 0.22 \ \mathring{A}$ and $	heta = 60^{\circ}$.Substituting the values:$0.22 - \lambda_1 = 0.024(1 - \cos 60^{\circ})$$0.22 - \lambda_1 = 0.024(1 - 0.5)$$0.22 - \lambda_1 = 0.024 	imes 0.5$$0.22 - \lambda_1 = 0.012$$\lambda_1 = 0.22 - 0.012 = 0.208 \ \mathring{A}$.

In the Compton scattering process,the incident $X$-radiation is scattered at an angle of $60^{\circ}$. The wavelength of the scattered radiation is $0.22 \ \mathring{A}$. The wavelength of the incident $X$-radiation in $\mathring{A}$ units is:

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