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. E_max) of emitted electron is given by the following equation

K. E_max = hν – W_{0 ………} 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

⇒

⇒ KE_max = 0.34 eV

Or in Joules,

KE_max = 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.E_max

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.E_max = 1/2 mv²_max

m = mass of the electron = 9.1 × 10^-31 kg

v_max = maximum velocity of electron

v²_max = 2 K.E_max / m

v_max = (2 × 0.546 × 10^-19 J / 9.1 × 10^-31)^1/2

= 347 km /sec

Maximum velocity of emitted electron is 347 km / sec.