Chapter 27, Problem 21P

To determine

(a) To determine:

Work function of the material.

Answer to Problem 21P

Solution:

Work function of the material is $2.25\text{eV}$.

Explanation of Solution

Given:

Minimum wavelength, $\lambda =550\text{nm}$

Formula used:

The work function is calculated by the formula

$W=\frac{hc}{e\lambda }\left(\text{eV}\right)$

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 work function is calculated by the formula

$W=\frac{hc}{e\lambda }\left(\text{eV}\right)$

Plugging the values in the above equation

$\begin{array}{c}W=\frac{6.62\times {10}^{-34}\times 3\times {10}^8}{1.6\times {10}^{-19}\times 550\times {10}^{-9}}\\ =2.25\text{eV}\end{array}$

To determine

(b) To determine:

Stopping voltage

Answer to Problem 21P

Solution:

The stopping voltage required is $0.85\text{V}$.

Explanation of Solution

Given:

Wavelength, $\lambda =400\text{nm}$

Formula used:

The 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 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 $2.25\text{eV}$, so the excess energy of the electron is

$\begin{array}{c}E=W-W_f\\ =3.10-2.25\\ =0.85\text{eV}\end{array}$

Now the stopping voltage required is calculated as

$\begin{array}{c}E=eV\\ 0.85e=eV\\ V=0.85\text{volt}\end{array}$