The proton and α - particle are accelerated through same potential difference. Then the ratio of the de-Broglie wavelength of proton and α - particle is (mass of α -particle is 4 times mass of proton, charge of α -particle is 2 times charge of proton) (a) 3√(3) (b) 3√(2) (c) 2√(3) (d) 2√(2)

Answer: (d). λ = (h)/(√(2mqV)) ⇒ λ ∝ (1)/(√(mq)) for the same V. (λp)/(λα) = √((mα qα)/(mp qp)) = √(4×2) = 2√(2). So the answer is (d) 2√(2).

The proton and α - particle are accelerated through same potential… | MHT CET 2026 (20 Apr, Shift 1)

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Matter Waves: de Broglie Wavelength

The proton and $\alpha$ - particle are accelerated through same potential difference. Then the ratio of the de-Broglie wavelength of proton and $\alpha$ - particle is (mass of $\alpha$ -particle is $4$ times mass of proton, charge of $\alpha$ -particle is $2$ times charge of proton)

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Answer: (d)

$\lambda = \dfrac{h}{\sqrt{2mqV}} \Rightarrow \lambda \propto \dfrac{1}{\sqrt{mq}}$ for the same $V$ .
$\dfrac{\lambda_p}{\lambda_\alpha} = \sqrt{\dfrac{m_\alpha q_\alpha}{m_p q_p}} = \sqrt{4\times2} = 2\sqrt{2}$ .
So the answer is (d) $2\sqrt{2}$ .

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Matter Waves: de Broglie Wavelength question from MHT CET 2026 (20 Apr, Shift 1)

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