What is the de Broglie wavelength of an electron with a kinetic energy of 120 eV?
(Given: h=6.63×10−34Js,me=9.11×10−31kg,e=1.6×10−19coulomb
A
725pm
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B
500pm
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C
322pm
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D
112pm
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Solution
The correct option is D112pm For a particle, de Broglie wavelength λ=hp
Kinetic energy K=p22m ∴λ=h√2mK
For an electron λ=h√2meKe λ=(6.63×10−34J)√2×(9.11×10−31kg)×(120×1.6×10−9J) λ=112×10−12m=112 pm[∴1pm=10−12m]