Light of wavelength 310 nm falls on a metal of work function 2.5 eV. The stopping potential is (take \(hc=1240\) eV nm)
A. 2.5 V
B. 1.0 V
C. 1.5 V
D. 4.0 V
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Answer: C. 1.5 V
\(E=\frac{1240}{310}=4\) eV; \(eV_0=4-2.5=1.5\) eV, so \(V_0=1.5\) V.
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· 23 Oct 2026
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