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Question:

The electric field of light wave is given as E = 10⁻³ cos(2π × 5 × 10⁻⁷ - 2π × 6 × 10¹⁴ t) x̂ N/C. This light falls on a metal plate of work function 2eV. The stopping potential of the photo-electrons is: Given E (in eV) = 12375/λ (in Å)

0.48V

2.0V

2.48V

0.72V

Solution:

Correct option is A. 0.48V
ω = 6 × 10¹⁴ × 2π; f = 6 × 10¹⁴
C = fλ
λ = C/f = (3 × 10⁸)/(6 × 10¹⁴) = 5000 Å
Energy of photon → 12375/5000 = 2.475 eV
From Einstein's equation
KEmax = E - φ
eVs = 2.475 - 2;
eVs = 0.475 eV
Vs = 0.475 V
V = 0.48 volt