First slide

Matter waves (DC Broglie waves)

Question

If the kinetic energy of a free electron doubles, its de-Broglie wavelength changes by a factor

Easy

A

1 /2
B

2
C

$1 / \sqrt{2}$
D

3

Solution

we know that
$λ = \frac{h}{mv} and K = \frac{1}{2} {mv}^{2} K = \frac{(mv)^{2}}{2 m} or mv = \sqrt{2 mK} λ = \frac{h}{\sqrt{2 mK}} or λ \propto \frac{1}{\sqrt{K}} Now \frac{λ_{2}}{λ_{1}} = \frac{\sqrt{K_{1}}}{\sqrt{K_{2}}} = \frac{\sqrt{K_{1}}}{\sqrt{2 K_{1}}} (∵ K_{2} = 2 K_{1}) or \frac{λ_{2}}{λ_{1}} = \frac{1}{\sqrt{2}} or λ_{2} = \frac{λ_{1}}{\sqrt{2}}$

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