Question

A $\mathbf{0}\mathbf{.}\mathbf{065}\mathbf{}\mathit{n}\mathit{m}$X-ray source is directed at a sample of carbon. Determine the minimum speed of scattered electrons.

Short answer

The minimum speed of the scattered electrons is$1.12\times {10}^7\frac{m}{s}$.

Step-by-step solution

Identification of the given data:

The given data can be listed below as,

The wavelength of X-rays is $\lambda =0.065\hspace{0.33em}nm.$

Significance of wavelength of a photon:

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.

Determination of the minimum speed of the scattered electron

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

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

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 $9.1\times {10}^{-31}kg,\lambda$ is the wavelength, $h$ is the Plank’s constant whose value is$6.63\times {10}^{-34}\hspace{0.33em}J\cdot s.$.

Substitute all the known values in the above equation.

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

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