The ratio of the accelerating potentials required to accelerate (i)… | MHT CET 2026 (17 Apr, Shift 2)

Q11MHT CET 2026 · 17 APR · SHIFT 2Medium

Matter Waves: de Broglie Wavelength

The ratio of the accelerating potentials required to accelerate (i) an $\alpha$ -particle and (ii) a proton to have the same de Broglie wavelength associated with them is

(mass of $\alpha$ - particle = $6.4 \times 10^{-27}$ kg, mass of proton = $1.6 \times 10^{-27}$ kg)

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Answer: (a)

$\lambda = \dfrac{h}{\sqrt{2mqV}}$ ; same $\lambda$ needs the same $mqV$ .
$\dfrac{V_\alpha}{V_p} = \dfrac{m_p q_p}{m_\alpha q_\alpha} = \dfrac{1.6 \times 10^{-27} \times e}{6.4 \times 10^{-27} \times 2e} = \dfrac{1}{8}$
So the answer is (a) $1 : 8$ .

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