Question

if radiation of wavelength 5,000 angstrom is incident on a surface of work function 1.2 electron volt find the value of stopping potential at​

Answer 1

Dear Student,

● Answer -

V = 1.286 V

◆ Explaination -

# Given -

λ = 5000 A° = 5×10^-7 m

Φ = 1.2 eV = 1.2×1.6×10^-19 J

# Solution -

By Einstein's photoelectric eqn,

hν = Φ + KE

hc/λ = Φ + eV

eV = hc/λ - Φ

V = hc/eλ - Φ/e

V = (6.63×10^-34 × 3×10^8) / (1.6×10^-19 × 5×10^-7) - 1.2×1.6×10^-19 / 1.6×10^-19

V = 2.486 - 1.2

V = 1.286 V

Therefore, stopping potential is 1.286 V.

Thanks dear...