The work function of caesium metal is 2.14 eV.

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

 

Hence, 332.3 km/s is the maximum speed of the emitted photoelectrons.