JEE Main 2014 · 12 Apr · Shift 1 · Physics Q26

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PhysicsSingle correct+4 / -1Question 26 of 90

A beam of light has two wavelengths 4972 Å and 6216 Å with a total intensity of $3.6\times10^{-3}$ Wm $^{-2}$ equally distributed among the two wavelengths. The beam falls normally on an area of 1 cm $^2$ of a clean metallic surface of work function 2.3 eV. Assume that there is no loss of light by reflection and that each capable photon ejects one electron. The number of photo electrons liberated in 2s is approximately :

6\times10^{11}

9\times10^{11}

11\times10^{11}

15\times10^{11}

Answer and worked solution

Official answer: B

Photon energies: $\frac{12420}{4972}=2.50$ eV (above 2.3 eV) and $\frac{12420}{6216}=2.00$ eV (below). Only the 4972 Å half ejects electrons.

Energy in 2 s on 1 cm²: $1.8\times10^{-3}\times10^{-4}\times2=3.6\times10^{-7}$ J.
$N=\frac{3.6\times10^{-7}}{2.5\times1.6\times10^{-19}}\approx9\times10^{11}.$

Topics:Photoelectric effect

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JEE Main 2014 · 12 Apr · Shift 1, Physics question 26