Chapter 27, Problem 22P

To determine

Maximum kinetic energy of electrons.

Answer to Problem 22P

Solution:

Maximum kinetic energy of electrons is $0.62\text{eV}$.

Explanation of Solution

Given:

Work function of barium, $W_o=2.48\text{eV}$

Minimum wavelength of visible spectrum, $\lambda =400\text{nm}=\text{400}\times {\text{10}}^{-9}\text{m}$

Formula used:

The maximum energy of the photon is calculated as

$W=\frac{hc}{e\lambda }$

Here,

$e$ is the charge of electron

$h=6.62\times {10}^{-34}\text{J-s}$ is the Planck’s constant

$c=3\times {10}^8\text{m/s}$ is the speed of light

$\lambda$ is the wavelength

Calculation:

The maximum energy of the photon is calculated as

$W=\frac{hc}{e\lambda }$

Plugging the values in the above equation

$\begin{array}{c}W=\frac{6.62\times {10}^{-34}\times 3\times {10}^8}{1.6\times {10}^{-19}\times 400\times {10}^{-9}}\\ =3.10\text{eV}\end{array}$

Since the work function of the metal is $2.48\text{eV}$, so the kinetic energy of the electron is

$\begin{array}{c}E=W-W_o\\ =3.10-2.48\\ =0.62\text{eV}\end{array}$