X-rays of wavelength 0.140 ,nm are scattered | vedclass

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

Vedclass · Accepted Answer

$0.142$

Vedclass · Answer

(B) The Compton shift formula is given by $\Delta \lambda = \lambda^{\prime} - \lambda = \frac{h}{m_e c} (1 - \cos \phi)$.
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 is $\frac{h}{m_e c} \approx 2.426 	imes 10^{-12} \,m = 0.002426 \,nm$.
Thus,$\lambda^{\prime} = \lambda + \Delta \lambda = 0.140 \,nm + 0.002426 \,nm = 0.142426 \,nm$.
Rounding to three decimal places,we get $\lambda^{\prime} \approx 0.142 \,nm$.

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

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