Chapter 27, Problem 23P

To determine

(a) To determine:

Maximum kinetic energy of electrons.

Answer to Problem 23P

Solution:

Maximum kinetic energy of electrons is $0.92\text{eV}$.

Explanation of Solution

Given:

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

Minimum wavelength of UV light, $\lambda =365\text{nm}=365\times {\text{10}}^{-9}\text{m}$

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

$\begin{array}{c}W=\frac{6.62\times {10}^{-34}\times 3\times {10}^8}{1.6\times {10}^{-19}\times 365\times {10}^{-9}}\\ =3.40\text{eV}\end{array}$

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

$\begin{array}{c}KE=W-W_o\\ =3.40-2.48\\ =0.92\text{eV}\end{array}$

To determine

(b) To determine:

Maximum speed of the electron

Answer to Problem 23P

Solution:

Maximum speed of the electron is $5.7\times {10}^5\text{m/s}$.

Explanation of Solution

Given:

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

Minimum wavelength of UV light, $\lambda =365\text{nm}=365\times {\text{10}}^{-9}\text{m}$

Formula used:

The kinetic energy is given by the formula

$KE=\frac{1}{2}m{v}^2$

Here,

$m$ is the mass of the electron

$v$ is the velocity of the electron

Calculation:

The velocity of the electron is calculated as

$KE=\frac{1}{2}m{v}^2$

Plugging the values in the above equation

$\begin{array}{c}0.92\times 1.6\times {10}^{-19}=\frac{1}{2}\times 9.1\times {10}^{-31}\times v^2\\ v=5.7\times {10}^5\text{m/s}\end{array}$