If the kinetic energy of a free electron doubles, it's de Broglie wavelength ( λ) changes by a factor (a) 1/√(2) (b) 1/2 (c) √(2) (d) 2

Answer: (a). λ = (h)/(√(2mK)) ⇒ λ ∝ 1/√(K). K' = 2K ⇒ λ' = (λ)/(√(2)). So the answer is (a) 1/√(2).

If the kinetic energy of a free electron doubles, it's de Broglie… | MHT CET 2022 (11 Aug, Shift 2)

Q79MHT CET 2022 · 11 AUG · SHIFT 2Easy

Matter Waves: de Broglie Wavelength

If the kinetic energy of a free electron doubles, it's de Broglie wavelength ( $\lambda$ ) changes by a factor

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Answer: (a)

$\lambda = \dfrac{h}{\sqrt{2mK}} \Rightarrow \lambda \propto \dfrac{1}{\sqrt{K}}$ .
$K' = 2K \Rightarrow \lambda' = \dfrac{\lambda}{\sqrt{2}}$ .
So the answer is (a) $\dfrac{1}{\sqrt{2}}$ .

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