Question

U. V. light of wavelength 200 nm is incident on polished surface of Fe. Work function of the surface is 4.5 eV. Find maximum speed of phote electrons.

(h = 6.625 × 10^–34 J.s, c = 3 × 10⁸ ms^–1, 1 eV = 1.6 × 10^–19 J)

(A) 7.75 × 10⁴ (m/s)

(B) 875 × 10⁵(m/s)

(C) 8.75 × 10⁴(m/s)

(D) 7.75 × 10⁵ (m/s)

Answer 1

The correct option is (D) 7.75 × 10⁵ (m/s).

Explanation:

We know KE = eV = (hc / λ) – ɸ

∴ V = (hc / eλ) – (ɸ/e)

∴ stopping potential = V = [{6.625 × 10^–34 × 3 × 10⁸} / {1.6 × 10^–19}] – 4.5(in eV)

= 6.21 – 4.5 = 1.71 V

Maximum Kinetic energy = eV

∴ (1/2)mV_max² = eV

V_max = √{(2eV) / m}

= √[(2 × 1.6 × 10^–19 × 1.71) / (9.11 × 10^–31)]

∴ V_max = 7.75 × 10⁵ m/s