 # 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, and

(c) maximum speed of the emitted photoelectrons?

Solution:

Work function of caesium metal, $\phi_{0}=2.14 \mathrm{eV}$

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

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

$K=h v-\phi_{0}$

Where,

h = Planck’s constant = 6.626 × 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, the maximum kinetic energy of the emitted electrons is 0.345 eV.

(b)For stopping potential $V_{0}$, we can write the equation for kinetic energy as:

$K=e V_{0}$

$\therefore V_{0}=\frac{K}{e}$

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

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

(c)Maximum speed of the emitted photoelectrons = v

Hence, the relation for kinetic energy can be written as:

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

Where,

m = Mass of an electron = 9.1 × 10−31 kg

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

$=\frac{2 \times 0.345 \times 1.6 \times 10^{-19}}{9.1 \times 10^{-31}}=0.1104 \times 10^{12}$

$\therefore v=3.323 \times 10^{5} \mathrm{~m} / \mathrm{s}=332.3 \mathrm{~km} / \mathrm{s}$

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