de-Broglie wavelength associated with an electron accelerated through a potential difference ' V ' is ' λ '. When the accelerating potential is increased to ' 4 V ', de-Broglie wavelength. (a) reduces to half (b) remains the same (c) reduces to (1/4) th (d) increases by 25%

Answer: (a). λ ∝ 1/√(V). V' = 4V ⇒ λ' = (λ)/(2). So the answer is (a) reduces to half.

de-Broglie wavelength associated with an electron accelerated through… | MHT CET 2021 (23 Sep, Shift 2)

Q90MHT CET 2021 · 23 SEP · SHIFT 2Easy

Matter Waves: de Broglie Wavelength

de-Broglie wavelength associated with an electron accelerated through a potential difference ' $\mathrm{V}$ ' is ' $\lambda$ '. When the accelerating potential is increased to ' $4\ \mathrm{V}$ ', de-Broglie wavelength.

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Answer: (a)

$\lambda \propto \dfrac{1}{\sqrt{V}}$ .
$V' = 4V \Rightarrow \lambda' = \dfrac{\lambda}{2}$ .
So the answer is (a) reduces to half.

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