Question

Determine the wavelength of an X-ray photon that can impart, at most $\mathbf{80}\mathbf{}\mathit{k}\mathit{e}\mathit{V}$ of kinetic energy to a free electron.

Short answer

The wavelength of an X-ray photon is$\mathbf{1}\mathbf{.}\mathbf{55}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{11}}\mathbf{\hspace{0.33em}}\mathit{m}\mathbf{.}$

Step-by-step solution

Identification of the given data:

The given data can be listed below as,

The kinetic energy of the photon is $E=80keV.$.

Significance of kinetic energy of an electron:

The kinetic energy of an X-ray photon can be obtained with the help of the energy equation. The relation between kinetic energy and a photon’s wavelength is a direct linear one.

Determination of the wavelength of an X-ray photon

The relation of the wavelength of an X-ray photonis expressed as,

$\lambda =\frac{hc}{E}$

Here,$\lambda$ is thewavelength of an X-ray photon, $h$is the Plank’s constant whose value is $6.63\times {10}^{-34}\hspace{0.33em}J\cdot s$and $c$is the light speed in vacuum whose value is$3\times {10}^8\hspace{0.33em}m/s$.

Substitute all the known values in the above equation.

localid="1657611613977" $$$$\begin{gathered} \lambda =\left[\frac{\left(6.63\times {10}^{-34}\mathrm{J}·\mathrm{s}\right)\left(3\times {10}^8\mathrm{m}/\mathrm{s}\right)}{\left(80\mathrm{keV}\right)\times \left(\frac{1.6\times {10}^{-16}\mathrm{J}}{1\mathrm{keV}}\right)}\right] \\ =1.55\times {10}^{-11}m \end{gathered}$$

Thus, the wavelength of an X-ray photonis$1.55\times {10}^{-11}m$.