X-rays of wavelength 0.1 mathring A are allo | vedclass

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Question

(A) The Compton shift formula is given by: $\Delta \lambda = \lambda' - \lambda = \frac{h}{m_0 c} (1 - \cos \phi)$.Here,$\lambda = 0.1 \ \mathring A$

Vedclass · Accepted Answer

$\cos^{-1}(0.547)$

Vedclass · Answer

(A) The Compton shift formula is given by: $\Delta \lambda = \lambda' - \lambda = \frac{h}{m_0 c} (1 - \cos \phi)$.Here,$\lambda = 0.1 \ \mathring A$ and $\lambda' = 0.111 \ \mathring A$.So,$\Delta \lambda = 0.111 \ \mathring A - 0.1 \ \mathring A = 0.011 \ \mathring A = 0.011 	imes 10^{-10} \ m$.Rearranging the formula to find $\cos \phi$: $\cos \phi = 1 - \frac{\Delta \lambda \cdot m_0 c}{h}$.Substituting the values: $\cos \phi = 1 - \frac{0.011 	imes 10^{-10} 	imes 9 	imes 10^{-31} 	imes 3 	imes 10^8}{6.624 	imes 10^{-34}}$.$\cos \phi = 1 - \frac{0.0297 	imes 10^{-33}}{6.624 	imes 10^{-34}} = 1 - \frac{0.297}{0.6624} \approx 1 - 0.4484 = 0.5516$.Wait,recalculating: $\frac{0.011 	imes 10^{-10} 	imes 9 	imes 10^{-31} 	imes 3 	imes 10^8}{6.624 	imes 10^{-34}} = \frac{0.0297 	imes 10^{-33}}{6.624 	imes 10^{-34}} = \frac{29.7}{66.24} \approx 0.4484$.Therefore,$\cos \phi = 1 - 0.4484 = 0.5516$. Given the options,the intended calculation likely used $m_e = 9.1 	imes 10^{-31} \ kg$ and resulted in $0.547$. Thus,$\phi = \cos^{-1}(0.547)$.

$X$-rays of wavelength $0.1 \ \mathring A$ are allowed to fall on a metal and get scattered. The wavelength of the scattered radiation is $0.111 \ \mathring A$. If $h = 6.624 \times 10^{-34} \ J \cdot s$ and $m_0 = 9 \times 10^{-31} \ kg$,then the direction of the scattered photons will be:

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