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JEE Main 2022 · 26 Jun · Shift 2, Physics question 60

The stopping potential for photoelectrons emitted from a surface illuminated by light of wavelength 6630 Å is 0.42 V. If the threshold frequency is x×1013x\times10^{13} /s, where xx is ______ (nearest integer).

(Given, speed light =3×108=3\times10^8 m/s, Planck's constant =6.63×10−34=6.63\times10^{-34} Js)

Answer and worked solution

Official answer: 35

Photon energy. hcλ=6.63×10−34×3×1086.63×10−7=3×10−19\dfrac{hc}\lambda=\dfrac{6.63\times10^{-34}\times3\times10^8}{6.63\times10^{-7}}=3\times10^{-19} J =1.875=1.875 eV.

Work function. 1.875−0.42=1.4551.875-0.42=1.455 eV ≈2.33×10−19\approx2.33\times10^{-19} J.

Threshold. ν0=2.33×10−196.63×10−34≈3.5×1014=35×1013\nu_0=\dfrac{2.33\times10^{-19}}{6.63\times10^{-34}}\approx3.5\times10^{14}=35\times10^{13} Hz.

Answer: 35

Topics:Photoelectric effect
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