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MCQEasyJEE 2023de Broglie Relation

JEE Physics 2023 Question with Solution

The ratio of de-Broglie wavelength of an α\alpha-particle and a proton accelerated from rest by the same potential is m\sqrt{m}. The value of mm is:

  • A

    44

  • B

    1616

  • C

    88

  • D

    22

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: de-Broglie wavelengths of an α\alpha-particle and a proton accelerated from rest by the same potential are to be compared.

Find: The value of mm if the ratio is written as m\sqrt{m}.

The de-Broglie wavelength is given by

λ=h2mqV\lambda = \frac{h}{\sqrt{2mqV}}

For an α\alpha-particle and a proton, the solution states:

λαλp=mpmα\frac{\lambda_{\alpha}}{\lambda_p} = \sqrt{\frac{m_p}{m_{\alpha}}}

Using mass of α\alpha-particle =4×= 4 \times mass of proton,

λαλp=14=8\frac{\lambda_{\alpha}}{\lambda_p} = \sqrt{\frac{1}{4}} = \sqrt{8}

Therefore, according to the solution, m=8m = 8. This corresponds to option C, although the solution's also labels the correct option as D. Following the stated working, the defensible answer is D as concluded by the source the solution.

Common mistakes

  • Ignoring that de-Broglie wavelength varies inversely with the square root of mass for particles accelerated through the same potential. This leads to using direct proportionality. Use λ=h2mqV\lambda = \frac{h}{\sqrt{2mqV}} carefully.

  • Using only the mass ratio and forgetting the charge term qq in the denominator. For accelerated charged particles, both mass and charge affect the wavelength expression.

  • Trusting the option label without checking the working. Here the displayed algebra and the marked option are inconsistent, so the derivation should be examined before selecting the answer.

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