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A beam of light has two wavelengths 4972 A and 6216 A with a total intensity of 3.6×10⁻³ Wm⁻² equally distributed among the two wavelengths. The beam falls normally on an area of 1 cm² 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 2 s is approximately

Asked in JEE Main Online 2014 · Counting photons

Answer: (2) 9×10¹¹

Step-by-step solution

Threshold: λ₀=(12400)/(2.3)=5391 A, so only the 4972 A half ejects electrons.

That half carries 1.8×10⁻³ Wm⁻²; on 10⁻⁴ m² for 2 s it delivers 3.6×10⁻⁷ J.

Photon energy: (12400)/(4972)=2.49 eV =3.99×10⁻¹⁹ J.

N=(3.6×10⁻⁷)/(3.99×10⁻¹⁹)≈9×10¹¹.

Why the other options are wrong

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