An electron accelerated by a potential difference ' V ' has de-Broglie wavelength ' λ '. If the electron is accelerated by a potential difference ' 9 V ', its de-Broglie wavelength will be (a) (λ)/(4.5) (b) (λ)/(3) (c) (λ)/(2) (d) λ

Answer: (b). λ ∝ 1/√(V). V' = 9V ⇒ λ' = (λ)/(3). So the answer is (b) (λ)/(3).

An electron accelerated by a potential difference ' V ' has… | MHT CET 2025 (19 Apr, Shift 2)

Q31MHT CET 2025 · 19 APR · SHIFT 2Easy

Matter Waves: de Broglie Wavelength

An electron accelerated by a potential difference ' $V$ ' has de-Broglie wavelength ' $\lambda$ '. If the electron is accelerated by a potential difference ' 9 V ', its de-Broglie wavelength will be

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Answer: (b)

$\lambda \propto \dfrac{1}{\sqrt{V}}$ .
$V' = 9V \Rightarrow \lambda' = \dfrac{\lambda}{3}$ .
So the answer is (b) $\dfrac{\lambda}{3}$ .

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Matter Waves: de Broglie Wavelength question from MHT CET 2025 (19 Apr, Shift 2)

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