X rays with a wavelength of 71 pm are directed onto a gold foil and eject tightly bound electrons from the gold atoms. The ejected electrons then move in circular paths of radius r in a region of the uniform magnetic field $\overrightarrow{\mathbf{B}}$. For the fastest of the ejected electrons, the product Br is equal to localid="1664288408568" $\mathbf{1}\mathbf{.}\mathbf{88}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{T}\mathbf{·}\mathbf{m}$. Find (a) the maximum kinetic energy of those electrons and (b) the work done in removing them from the gold atoms.

The given data is listed below.

The product of B and r is $\mathrm{B}\times \mathrm{r}=1.88\times {10}^{-4}\mathrm{T}.\mathrm{m}$

The wavelength of the X rays is$\lambda =71pm$

The kinetic energy of a particle equals the amount of work required to accelerate the particle from rest to speed v. Therefore, kinetic energy on the particle is given by-

$\mathrm{K}=\frac{1}{2}\mathrm{m}{v}^2$

The kinetic energy is a scalar, always positive or zero.

The radius is given by

$r=\frac{m_{e}v}{qB}$ ….. (1)

Here, $m_e$is the mass of the electron, q is the charge of the electron, v is the speed of an electron, and B is the magnetic field.

Now, the speed of an electron can be obtained from equation (1)

$v=\frac{rBq}{m_e}$

Therefore, kinetic energy is given by-

$\begin{array}{c}{K}_{\max }=\frac{1}{2}{m}_e{v}^2\\ =\frac{1}{2}{m}_e{\left(\frac{rBq}{m_e}\right)}^2\\ =\frac{{\left(rB\right)}^2{q}^2}{2m_e}\end{array}$

Thus, the maximum kinetic energy of the electrons is 3.1 keV.

Write the equation for energy of photon as below

$E_{photon}=\frac{hc}{\lambda }$

Here,

The Plank’s constant is

$$\begin{gathered} \mathrm{h}=6.626\times {10}^{-34}{\mathrm{m}}^2.\mathrm{kg}/\mathrm{s} \\ =\frac{6.626\times {10}^{-34}{\mathrm{m}}^2.\mathrm{kg}/\mathrm{s}}{1.6\times {10}^{-19}\mathrm{J}} \\ =4.141\times {10}^{-15}\mathrm{eV}.\mathrm{s} \end{gathered}$$

The speed of light is,

$$\begin{gathered} \mathrm{c}=3\times {10}^8\mathrm{m}/\mathrm{s} \\ =3\times {10}^{17}\mathrm{nm}/\mathrm{s} \end{gathered}$$

Therefore, the value of hc is,

Hence, the work done in removing the electrons from gold atoms is 14 keV.