The ratio of wavelengths of proton and deuteron accelerated by potential Vp and Vd is 1: √(2). Then the ratio of Vp to Vd will be: (a) 1: 1 (b) √(2): 1 (c) 2: 1 (d) 4: 1

Answer: (d). λ = (h)/(√(2mqV)); proton and deuteron have the same charge, md = 2mp. (λp)/(λd) = √((mdVd)/(mpVp)) = √((2Vd)/(Vp)) = 1/√(2) (2Vd)/(Vp) = 1/2 ⇒ (Vp)/(Vd) = 4 So the answer is (d) 4:1.

The ratio of wavelengths of proton and deuteron accelerated by… | JEE Main 2022 (25 Jul, Shift 2)

Q85JEE Main 2022 · 25 JUL · SHIFT 2Medium

Matter Waves: de Broglie Wavelength

The ratio of wavelengths of proton and deuteron accelerated by potential Vp and Vd is 1: $\sqrt{2}$ . Then the ratio of Vp to Vd will be:

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Answer: (d)

$\lambda = \dfrac{h}{\sqrt{2mqV}}$ ; proton and deuteron have the same charge, $m_d = 2m_p$ .
$\dfrac{\lambda_p}{\lambda_d} = \sqrt{\dfrac{m_dV_d}{m_pV_p}} = \sqrt{\dfrac{2V_d}{V_p}} = \dfrac{1}{\sqrt{2}}$
$\dfrac{2V_d}{V_p} = \dfrac{1}{2} \Rightarrow \dfrac{V_p}{V_d} = 4$
So the answer is (d) $4:1$ .

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