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JEE Main 2026 · 5 Apr · Shift 1, Physics question 41

Light source having wavelength 331 nm is used to generate photo-electrons whose stopping potential is 0.2 V. The work function of the used metal in the experiment is α×10−19\alpha\times10^{-19} J. The value of α\alpha is ________.
(h=6.62×10−34h=6.62\times10^{-34} J s, e=1.6×10−19e=1.6\times10^{-19} C and c=3×108c=3\times10^8 m/s)

Choose one option
Answer and worked solution

Official answer: C

Photon energy.
E=hcλ=6.62×10−34×3×108331×10−9=6.00×10−19 J.E=\frac{hc}\lambda=\frac{6.62\times10^{-34}\times3\times10^8}{331\times10^{-9}}=6.00\times10^{-19}\text{ J}.

Kinetic energy. eV0=0.32×10−19eV_0=0.32\times10^{-19} J.

Work function: 6.00−0.32=5.686.00-0.32=5.68, so α=5.68\alpha=5.68.

(C)

Topics:Photoelectric effect
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