Question

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?

Answer 1

Hey Dear,

◆ Answers -

(a) KEmax = 5.54×10^-20 J

(b) Vo = 0.3463 V

(c) vmax = 3.49×10^5 m/s

● Explaination -

# Given -

Φo = 2.14 eV = 3.424×10^-19 V

ν = 6×10^14 Hz

# Solution -

(a) Maximum kinetic energy of emitted electrons -

KEmax = h.ν - Φo

KEmax = 6.63×10^-34 × 6×10^14 - 3.424×10^-19

KEmax = 2.485 - 2.14

KEmax = 5.54×10^-20 J

(b) Stopping potential is calculated as -

Vo = KEmax / e

Vo = 5.54×10^-20 / 1.6×10^-19

Vo = 0.3463 V

(c) Maximum speed of the emitted electrons is -

vmax = √(2.KEmax/m)

vmax = √(2 × 5.54×10^-20 / 9.1×10^-31)

vmax = 3.49×10^5 m/s

Hope this helps...