Question

The work function of caesium metal is 2.14 eV. When light of frequency 6 ×1014Hz 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?

Answer 1

Given: The work function of caesium metal is 2.14 eV and the frequency of light is 6× 10 14  Hz.

a)

The maximum kinetic energy (eV) in photoelectric effect is given as,

K.E.= hν e −ϕ

Where, the Plank’s constant is h, the charge on electron is e, the work function of caesium metal is ϕ ad the frequency of light is ν.

By substituting the given values in above equation, we get

K.E.= 6.626× 10 −34 ×6× 10 14 1.6× 10 −19 −2.14 =2.485−2.14 =0.345 eV

Thus, the maximum kinetic energy of emitted electron is 0.345 eV.

b)

The stopping potential is given as,

V 0 = K.E. e

By substitute the values in above equation, we get

V 0 = 0.345×1.6× 10 −19 1.6× 10 −19 =0.345 V

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

c)

The maximum speed of the photoelectrons is given as,

K.E.= 1 2 m v 2 v= 2×K.E. m

Where, the maximum speed of electron is v and the mass of electron is m.

By substitute the given values in above equation, we get

v= 2×0.345×1.6× 10 −19 9.1× 10 −31 =3.483× 10 5  m/s =348.3 km/s

Thus, the maximum speed of emitted photoelectron is 348.3 km/s.