An electron of mass m, when accelerated through a potential difference V, has de Broglie wavelength λ. The de Broglie wavelength associated with a proton of mass M accelerated through the same potential difference, will be (a) λ M/m (b) λ m/M (c) λ√(M/m) (d) λ√(m/M)

Answer: (d). For a charge q accelerated through V: λ = (h)/(√(2mqV)). Electron and proton have the same |q| and the same V, so λ ∝ 1/√(m). (λp)/(λ) = √(m/M) ⇒ λp = λ√(m/M) So the answer is (d) λ√(m/M).

An electron of mass m, when accelerated through a potential… | NEET 1995 (AIPMT)

Q150NEET 1995 · AIPMTMedium

Matter Waves: de Broglie Wavelength

An electron of mass $m$ , when accelerated through a potential difference $V$ , has de Broglie wavelength $\lambda$ . The de Broglie wavelength associated with a proton of mass $M$ accelerated through the same potential difference, will be

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Answer: (d)

For a charge $q$ accelerated through $V$ : $\lambda = \dfrac{h}{\sqrt{2mqV}}$ .
Electron and proton have the same $|q|$ and the same $V$ , so $\lambda \propto \dfrac{1}{\sqrt{m}}$ .
$\dfrac{\lambda_p}{\lambda} = \sqrt{\dfrac{m}{M}} \Rightarrow \lambda_p = \lambda\sqrt{\dfrac{m}{M}}$
So the answer is (d) $\lambda\sqrt{\frac{m}{M}}$ .

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