When a light of a given wavelength falls on a metallic surface the stopping potential for photoelectrons is 3.2 V. If a second light having wavelength twice of first light is used, the stopping potential drops to 0.7 V.

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Correct answer: A. 2.5 × 10^ -7

Using Einstein's photoelectric equation with stopping potentials:
For first light,

h\nu_1 = \phi + eV_{s1}

.
For second light with

\lambda_2 = 2\lambda_1

(so

\nu_2 = \nu_1/2

):

h\nu_1/2 = \phi + eV_{s2}

.
From given values:

h\nu_1 - 3.2e = h\nu_1/2 - 0.7e

gives

h\nu_1/2 = 2.5e

.
Thus

\lambda_1 = \frac{hc}{5e} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{5 \times 1.6 \times 10^{-19}} \approx 2.5 \times 10^{-7}

m.

When a light of a given wavelength falls on a metallic surface the...

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