Question

32.) Ultraviolet light by wavelength 200 nm is incident on polished surface
of Fe (Iron). Work function of the surface is 4.71 V. What will be its stopping
potential ?
(h = 6.626 x 10 9 Je: 16V - 1.6 10-19 J: C-3% 109 m)
(A) 1.5 V
(B) 2.5 V
(C) 0.6 V
(D) None of these​

Answer 1

Answer:

THE CORRECT OPTION IS B. 2.5 V

Explanation:

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Answer 2

Answer:

Explanation:

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)mVmax2 = eV Vmax = √{(2eV) / m} = √[(2 × 1.6 × 10–19 × 1.71) / (9.11 × 10–31)]

∴ Vmax = 7.75 × 105 m

light-wavelength-200-incident-on-polished-surface-of-fe-work-function-of-the-surface-is-ev

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