Question

The work function of a metal surface is 1.2ev. Calculate its kinetic energy when light of frequency 5.5*10^14 Hz falls on the surface

Answer 1

Answer:

Work function of caesium metal, ø0 = 2.14 eV

Frequency of light, v = 6.0 x 1014 Hz

(a) The maximum kinetic energy is given by the photoelectric effect as: K = hv - ø0

Where,

h = Planck’s constant = 6.626 × 10−34 Js

∴ K = 6.626 x 1034 x 6 x 1014 / 1.6 x 10-19 - 2.14 = 2.485 - 2.14 = 0.345 eV

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

( b ) For stopping potential , V0 we can write the equation for kinetic energy as: K = eV0

∴ V0 = K/e = 0.345 x 1.6 x 10-19 / 1.6 x 10-19 = 0.345 V

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

(c) Maximum speed of the emitted photoelectrons = vHence, the relation for kinetic energy can be written as: K = 1/2 mv2

Where,

m = Mass of an electron = 9.1 × 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.