Question

X-rays of wavelength 10.0pm are scattered from a target.(a)Find the wavelength of x-rays scattered through 45°(b)Find the maximum wavelength present in the scattered x-rays.(c)Find the maximum kinetic energy of the recoil electrons.​

Answer 1

Explanation:

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Answer 2

Given:

(a)  the wavelength of x-rays scattered through 45°

X-rays of wavelength 10.0pm

$\lambda_{\mathrm{c}}=\text { Compton wavelength of recoil electron }=2.426 \mathrm{pm} \text { } \phi=45^{\circ} ; \lambda=10.0 \mathrm{pm}$

(b) the maximum wavelength present in the scattered x-rays.

$\begin{array}{l}\lambda^{\prime}=\lambda+\lambda_{\mathrm{c}}(1-\cos \phi)=\left[10.0+2.426\left(1-\cos 45^{0}\right)\right] \mathrm{pm}==[10.0+2.426(1-0.707)] \mathrm{pm} \\\lambda^{\prime}=[10.0+(2.426 \times 0.293)] \mathrm{pm}=10.71 \mathrm{pm}\end{array}$$\text { Maximum wavelength of scattered } \mathrm{x} \text {-ray }=\lambda^{\prime} \max =\lambda+\Delta \lambda_{\max }=\lambda+2 \lambda_{\mathrm{c}}$

$=[10.0+(2 \times 2.426)] \mathrm{pm}=14.9 \mathrm{pm}$

(c) the maximum kinetic energy of the recoil electrons

$\begin{array}{l}\mathrm{KE}_{\max }=\left(6.62 \times 10^{-34}\right) \mathrm{J}-\mathrm{s}\left(3 \times 10^{8}\right) \mathrm{ms}^{-1}\left[\frac{1}{10 p m}-\frac{1}{14.9 p m}\right] \\\mathrm{KE}_{\max }=\left(6.62 \times 10^{-34}\right) \mathrm{J}-\mathrm{s}\left(3 \times 10^{8}\right) \mathrm{ms}^{-1}\left[\frac{1}{10 \mathrm{~m}}-\frac{1}{14.9 \mathrm{~m}}\right] \times 10^{12}\end{array}$

$=6.54 \times 10^{-15} \text { Joule }=40.8 \mathrm{keV}$