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?

Work function of caesium metal, φ_O = 2.14 eV and

frequency of light, v = 6.0 x 10¹⁴ Hz

Ans (a).
The maximum kinetic energy is given by the photoelectric effect as: K = hv — φ_O

Where, h = Planck’s constant = 6.626 x 10⁻³⁴ Js

Therefore ,K = (6.626 x 10⁻³⁴) x ( 6.0 x 10¹⁴) /(1.6 x 10⁻¹⁹) -2.140

= 2.485 -2.140 = 0.345 eV

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

Ans (b).
For stopping potential V₀ we can write the equation for kinetic energy as:

K=e V₀

Therefore, V₀= K/e = (0.345 x 1.6 x10⁻¹⁹) /(1.6 x 10⁻¹⁹ )

Hence ,the stopping potential of the material is 0.345 V

Ans (c).
Maximum speed of the emitted photoelectrons = v

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

K = 1/2  x (mv²)

Where, m = Mass of an electron = 9.1 x 10^-31 kg

v² =2K /m

= 2 x (0.345 x 1.6 x 10⁻¹⁹ )/(9.1 x 10^-31) = 0.1104 x 10¹²

v=3.323 x 10⁵ m/s = 332.3 km/s

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