# The work function of caesium metal is 2.14 eV.

Question:

The work function of caesium metal is 2.14 eV. When light of frequency 6 × 1014 Hz is incident on the metal surface, photoemission of electrons occurs. What is the

(a) maximum kinetic energy of the emitted electrons

(b) Stopping potential

(c) maximum speed of the emitted photoelectrons?

Solution:

Work function of caesium, $\Phi_{o}=2.14 \mathrm{eV}$

Frequency of light, $\mathrm{v}=6.0 \times 10^{14} \mathrm{~Hz}$

(a) The maximum energy (kinetic) by the photoelectric effect:

$\mathrm{K}=\mathrm{hv}-\Phi_{o}$

h = Planck’s constant = 6.626 x 10-34 Js

Therefore,

$K=\frac{6.626 \times 10^{-34} \times 6 \times 10^{14}}{1.6 \times 10^{-19}}-2.14$

$=2.485-2.140=0.345 \mathrm{eV}$

Hence, 0.345 eV is the maximum kinetic energy of the emitted electrons.

(b) For stopping potential Vo, we can write the equation for kinetic energy as:

K = eVo

Therefore, $\mathrm{V}_{0}=\frac{K}{e}$

$=\frac{0.345 \times 1.6 \times 10^{-19}}{1.6 \times 10^{-19}}$

= 0.345 V

Hence, 0.345 V is the stopping potential of the material.

(c) Maximum speed of photoelectrons emitted = ν

Following is the kinetic energy relation:

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

Where,

m = mass of electron = 9.1 x 10-31 Kg

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

$=\frac{2 \times 0.345 \times 1.6 \times 10^{-19}}{9.1 \times 10^{-31}}$

=0.1104 x 1012

Therefore, ν = 3.323 x 105 m/s = 332.3 km/s

Hence332.3 km/s is the maximum speed of the emitted photoelectrons.