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Question

Photoelectrons are liberated by ultraviolet light of wavelength 3000˚A from a metallic surface for which the photoelectric threshold is 4000˚A. Calculate de Broglie wavelength of electrons emitted with maximum kinetic energy.

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Solution

Explanation:

From Einsteins photoelectric equation we get:

hcλ=hcλth+K.E

where,
λ=Wave length=3000Ao=3000×108cm(given)


λth=threshold wave length=4000Ao=4000×108cm(given)

h=Planck's constant=6.625×1034Js

c=Velocity of light=3×108m/s

K.E=hcλhcλth=hc[1λ1λth]=6.625×1034×3×108[13000×10814000×108]

K.E=1.6565×1019J

K.E=12mv2

12mv2=1.6565×1019J

multiplying above equation with m(mass of electron)

m=9.1×1031kg

m2v2=2×1.6565×1019×9.1×1031

mv=5.49×1025

de Broglie λ=hmv=6.625×10345.49×1025=1.2×109=1.2nm

Hence the correct answer is λ=1.2nm



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