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Radhika Batra Grade: 11
        Find the maximum magnitude of the linear momentum of a photoelectron emitted when light of wavelength 400 nm falls on a metal having work function 2.5 eV.
3 years ago

Answers : (1)

Aditi Chauhan
askIITians Faculty
396 Points
										Sol. Energy of photoelectron

⇒ ½ mv^2 = hc/λ – hv base 0 = 4.14 * 10^-15 * 3 *10^8/4 * 10^-7 – 2.5 ev = 0.605 ev.
We known KE = P2/2m ⇒ P^2 = 2m * KE.
P^2 = 2 * 9.1 * 10^-31 * 0.605 * 1.6 * 10^-19
P = 4.197 * 10^-25 kg – m/s
3 years ago
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