Find the de-Broglie wavelength of an electron with kinetic energy of 120 eV

112 pm

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95 pm

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124 pm

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102 pm

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Solution

The correct option is A
112 pm

$λ = \frac{h}{p}$

where, h=Plank's constant, P=Momentum

Also, $P = \sqrt{2 m k}$

$λ = \frac{h}{\sqrt{2 m k}}$

Putting values we get,

$λ \approx 112 p m$

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