The work function of caesium metal is 2.14 eV. When the 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?
(a) maximum energy (K. Emax) of emitted electron is given by the following equation
K. Emax = hν – W0 ……… equation no. 1
Where h = plank’s constant = 6.63 × 10-34 J. sec
ν = max. frequency of X-ray
W0 = work function (minimum energy required to emit an electron from neutral atom)
W0 = 2.14 eV
And, ν = 6 × 1014 Hz
From equation No. 1
⇒ 
⇒ KEmax = 0.34 eV
Or in Joules,
KEmax = 0.34 × 1.6 × 10-19 = 0.546 × 10-19 J
maximum KE of emitted electron is 0.546 × 10-19 J
(b) Stopping Potential is calculated by following relation
eV0 = K.Emax
Magnitude of charge of the electron ‘e’ = 1.6 × 10-19 C
V0 = Stopping Potential
⇒ 
⇒ V0 = 0.34 V
Stopping Potential of emitted electron is 0.34 V
(c) K.Emax = 1/2 mv2max
m = mass of the electron = 9.1 × 10-31 kg
vmax = maximum velocity of electron
v2max = 2 K.Emax / m
vmax = (2 × 0.546 × 10-19 J / 9.1 × 10-31)1/2
= 347 km /sec
Maximum velocity of emitted electron is 347 km / sec.