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MCQEasyJEE 2023de Broglie Relation

JEE Physics 2023 Question with Solution

Proton (PP) and electron (ee) will have same de-Broglie wavelength when the ratio of their momentum is (assume mp=1849 mem_p = 1849 \, m_e):

Options:

  • A

    1:431 : 43

  • B

    43:143 : 1

  • C

    1:18491 : 1849

  • D

    1:11 : 1

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: Proton (PP) and electron (ee) have the same de-Broglie wavelength.

Find: The ratio of their momenta.

Use the de-Broglie wavelength formula:

λ=hp\lambda = \frac{h}{p}

If the wavelengths are equal, then:

λp=λe\lambda_p = \lambda_e

So,

hpp=hpe\frac{h}{p_p} = \frac{h}{p_e}

Cancelling hh from both sides,

pppe=1\frac{p_p}{p_e} = 1

Therefore, the ratio of momentum is 1:11 : 1. The correct option is D.

Direct Comparison

Given: The de-Broglie wavelength of proton and electron is the same.

Find: The ratio pproton:pelectronp_{proton} : p_{electron}.

For each particle,

pelectron=hλp_{electron} = \frac{h}{\lambda}

and

pproton=hλp_{proton} = \frac{h}{\lambda}

Hence,

pprotonpelectron=hλhλ=1\frac{p_{proton}}{p_{electron}} = \frac{\frac{h}{\lambda}}{\frac{h}{\lambda}} = 1

Therefore, the linear momentum ratio between proton and electron is 1:11 : 1. The correct option is D.

Common mistakes

  • Using mass ratio to compare momentum directly is incorrect here because equal de-Broglie wavelength depends on pp through λ=hp\lambda = \frac{h}{p}, not directly on mass. First equate the wavelengths, then conclude the momenta are equal.

  • Confusing momentum with kinetic energy is wrong because particles with different masses can have the same momentum but different kinetic energies. Use the de-Broglie relation only with momentum.

  • Choosing the option from the displayed solution letter without checking the actual option mapping can be misleading because the source solution shows an inconsistent letter. Verify the numerical ratio from the worked steps and then map it to the correct option.

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