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JEE Main 2022 · 28 Jun · Shift 1, Chemistry question 82

If the work function of a metal is 6.63×10−196.63\times10^{-19}J, the maximum wavelength of the photon required to remove a photoelectron from the metal is ______ nm. (Nearest integer)

[Given : h = 6.63×10−346.63\times10^{-34}J s, and c = 3×1083\times10^8 m s−1^{-1}]

Answer and worked solution

Official answer: 300

Threshold wavelength. λ=hcW=6.63×10−34×3×1086.63×10−19=3×10−7\lambda=\dfrac{hc}W=\dfrac{6.63\times10^{-34}\times3\times10^8}{6.63\times10^{-19}}=3\times10^{-7} m =300=300 nm.

Answer: 300

Topics:Photoelectric effect and spectra
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