X-rays of wavelength 0.140 nm are scattered | vedclass

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Question

(B) The Compton shift formula is given by $\Delta \lambda = \lambda' - \lambda = \frac{h}{m_e c} (1 - \cos \phi)$.Given $\lambda = 0.140 \ nm = 0.140

Vedclass · Accepted Answer

$0.142$

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(B) The Compton shift formula is given by $\Delta \lambda = \lambda' - \lambda = \frac{h}{m_e c} (1 - \cos \phi)$.Given $\lambda = 0.140 \ nm = 0.140 	imes 10^{-9} \ m$ and $\phi = 90^{\circ}$.Since $\cos 90^{\circ} = 0$,the shift is $\Delta \lambda = \frac{h}{m_e c}$.The Compton wavelength $\frac{h}{m_e c} \approx 2.426 	imes 10^{-12} \ m = 0.002426 \ nm$.Therefore,$\lambda' = \lambda + \Delta \lambda = 0.140 \ nm + 0.002426 \ nm = 0.142426 \ nm$.Rounding to three decimal places,we get $\lambda' \approx 0.142 \ nm$.

$X$-rays of wavelength $0.140 \ nm$ are scattered from a block of carbon. What will be the wavelength of $X$-rays scattered at $90^{\circ}$ (in $nm$)?

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