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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. The de-Broglie wavelength of electrons emitted with maximum kinetic energy is:

A
1.2 nm
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B
3.215 nm
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C
7.28 ˚A
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D
1.65 ˚A
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Solution

The correct option is B 1.2 nm
Kinetic energy= Quantum energy - Threshold energy

K.E=6.625×1034×3×1083000×1086.625×1034×3×1084000×108=1.6565×1019J

12mv2=1.6565×1019

m2×v2=2×1.6565×1019×9.1×1031

mv=5.49×1025

λ=hmv

λ=6.625×10345.49×1025=1.2×109m

I.e 1.2 nm (1nm=109m)

Hence option A is correct.

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