X-rays of wavelength 22 , pm are scattered fr | vedclass

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(A) The Compton shift $\Delta \lambda$ is given by the formula: $\Delta \lambda = \frac{h}{m_0 c} (1 - \cos \phi)$.Here, the Compton wavelength $\lamb

Vedclass · Accepted Answer

$2.2$

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(A) The Compton shift $\Delta \lambda$ is given by the formula: $\Delta \lambda = \frac{h}{m_0 c} (1 - \cos \phi)$.Here, the Compton wavelength $\lambda_c = \frac{h}{m_0 c} \approx 2.43 	imes 10^{-12} \, m = 2.43 \, pm$.The scattering angle $\phi = 85^{\circ}$.Substituting the values: $\Delta \lambda = 2.43 	imes (1 - \cos 85^{\circ})$.Since $\cos 85^{\circ} \approx 0.087$, we have $\Delta \lambda = 2.43 	imes (1 - 0.087) = 2.43 	imes 0.913 \approx 2.218 \, pm$.Rounding to the nearest given option, the Compton shift is approximately $2.2 \, pm$.

$X$-rays of wavelength $22 \, pm$ are scattered from a carbon target at an angle of $85^{\circ}$ to the incident beam. The Compton shift for $X$-rays is ......... $pm$.

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