Dual Nature of Matter

JEE Main level, single correct, medium

97 of 200

Question 97 of 200

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. The wavelength of first light is ____\_\_\_\_ m. (h=6.63×10−34 J.s,e=1.6×10−19C,c=3×108 m/s)\left(\mathrm{h}=6.63 \times 10^{-34} \mathrm{~J}. \mathrm{s}, \mathrm{e}=1.6 \times 10^{-19} \mathrm{C}, \mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)

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Answer: (A) 2.5×10−72.5 \times 10^{-7}

Using Einstein's photoelectric equation with stopping potentials:
For first light, hν1=ϕ+eVs1h\nu_1 = \phi + eV_{s1}.
For second light with λ2=2λ1\lambda_2 = 2\lambda_1 (so ν2=ν1/2\nu_2 = \nu_1/2): hν1/2=ϕ+eVs2h\nu_1/2 = \phi + eV_{s2}.
From given values: hν1−3.2e=hν1/2−0.7eh\nu_1 - 3.2e = h\nu_1/2 - 0.7e gives hν1/2=2.5eh\nu_1/2 = 2.5e.
Thus λ1=hc5e=6.63×10−34×3×1085×1.6×10−19≈2.5×10−7\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.

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