Question

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.

Short answer

1. The maximum kinetic energy is 3.1 keV.

2. The work done is 14 keV.

Step-by-step solution

 Identification of the given data:

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$\lambda =71pm$

Significance of the kinetic energy of a particle:

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.

(a) To determine the maximum kinetic energy of the electrons:

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.

(b) To determine the work done in removing the electrons from gold atoms:

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.