Question

When a beam of mono-energetic electrons is directed at a tungsten target, X-rays are produced with wavelengths no shorter than$\mathbf{0}\mathbf{.}\mathbf{062}\mathbf{nm}$ . How fast are the electrons in the beam moving?

Short answer

The speed of the electrons is$1.18\times {10}^{7}m/s.$

Step-by-step solution

Identification of the given data

The wavelength of the X-rays is$\lambda =0.062\text{nm.}$

Significance of wavelength of a photon

The wavelengths of aphoton help to determine the speed of electrons in a beam of mono-energetic electrons by using the relation of momentum of the photon.

Step 3:Determination thespeed of electron in the beam

The relation of thespeed of an electron in a beamis mathematically expressed as:

$\begin{array}{c}v=\frac{p_{photon}}{m_e}\\ =\frac{\left(\frac{h}{\lambda }\right)}{m_e}.\end{array}$

Here,$v$is the speed ofan electron in the beam,$\lambda$ is the wavelength with the value$0.062nm,$$p_{photon}$is the momentum of the photon,$m_e$is the mass of an electron, whose value is$9.1\times {10}^{-31}\text{\hspace{0.17em}kg}$, $h$ is the Plank’s constant whose value is$6.63\times {10}^{-34}\text{\hspace{0.17em}J}\cdot \text{s}$.

Replaceall the known values in the above equation.

$\begin{array}{c}v=\frac{\left(\frac{6.63\times {10}^{-34}\text{\hspace{0.17em}J}\cdot \text{s}}{0.062\text{\hspace{0.17em}nm}\times \frac{{10}^{-9}\text{\hspace{0.17em}m}}{1\text{\hspace{0.17em}nm}}}\right)}{(9.1\times {10}^{-31}\text{\hspace{0.17em}kg})}\\ =1.18\times {10}^{7}J\cdot s/m\cdot kg\\ =(1.18\times {10}^{7}J\cdot s/m\cdot kg)\times \left(\frac{1m/s}{1J\cdot s/m\cdot kg}\right)\\ =1.18\times {10}^{7}m/s.\end{array}$

Thus, the speed ofelectron in the beamis$1.18\times {10}^{7}m/s.$