91Ó°ÊÓ

The given data can be listed below as,

The wavelength of X-rays is $\mathrm{λ}=0.065 nm.$

Whenever a specific source of X-ray generates an X-ray beam and the beam strikes a particular material then there would be a scattering of electrons takes place.

The relation of a minimum speed of the scattered electron is expressed as,

$$\begin{gathered} v_{\min }=\frac{P_{\text{phosar}}}{m_e} \\ =\frac{\left(\frac{\mathrm{h}}{\mathrm{λ}}\right)}{{\mathrm{m}}_{\mathrm{e}}} \end{gathered}$$

Here, localid="1657612674908" $v_{min}$is the minimum speed of a scattered electron, localid="1657612156210" $p_{photon}$is the momentum of the photon, $m_e$is the mass of an electron whose value is $9.1×{10}^{-31}kg,\mathrm{λ}$ is the wavelength, $h$ is the Plank’s constant whose value is$6.63×{10}^{-34} Jâ‹\dots s.$.

Substitute all the known values in the above equation.

localid="1657612488743" $$\begin{gathered} V_{\min }=\frac{\left(\frac{6.63×{10}^{-34}\mathrm{J}·\mathrm{s}}{0.065\mathrm{nm}×\frac{{10}^{-9}\mathrm{m}}{1\mathrm{nm}}}\right)}{\left(9.1×{10}^{-31}\mathrm{kg}\right)} \\ =1.12×{10}^7\mathrm{J}·\mathrm{s}/\mathrm{m}·\mathrm{kg} \\ =\left(1.12×{10}^7\mathrm{J}·\mathrm{s}/\mathrm{m}·\mathrm{kg}\right)×\left(\frac{1\mathrm{m}/\mathrm{s}}{1\mathrm{J}·\mathrm{s}/\mathrm{m}·\mathrm{kg}}\right) \\ =1.12×{10}^{7}m/s \end{gathered}$$

Thus, the minimum speed of the scattered electron is$1.12×{10}^{7}m/s.$