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MCQEasyJEE 2023de Broglie Relation

JEE Physics 2023 Question with Solution

The de Broglie wavelength of an electron having kinetic energy EE is λ\lambda. If the kinetic energy of the electron becomes E4\dfrac{E}{4}, then its de Broglie wavelength will be

  • A

    λ2\dfrac{\lambda}{2}

  • B

    2λ2\lambda

  • C

    λ2\dfrac{\lambda}{\sqrt{2}}

  • D

    2λ\sqrt{2}\lambda

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: The initial kinetic energy of the electron is EE and its de Broglie wavelength is λ\lambda.

Find: The new de Broglie wavelength when the kinetic energy becomes E4\dfrac{E}{4}.

The de Broglie relation is

λ=hp\lambda = \frac{h}{p}

For a non-relativistic electron,

p=2mEp = \sqrt{2mE}

So,

λ∝1E\lambda \propto \frac{1}{\sqrt{E}}

Initially, kinetic energy =E= E and wavelength =λ= \lambda.

When the new kinetic energy is

E4\frac{E}{4}

the new wavelength λ′\lambda' is

λ′∝1E/4=1E/2=2⋅1E\lambda' \propto \frac{1}{\sqrt{E/4}} = \frac{1}{\sqrt{E}/2} = 2\cdot \frac{1}{\sqrt{E}}

Hence,

λ′=2λ\lambda' = 2\lambda

Therefore, the de Broglie wavelength becomes 2λ2\lambda. The correct option is B.

Direct Proportionality Trick

Given: λ∝1E\lambda \propto \dfrac{1}{\sqrt{E}}.

Find: How wavelength changes when energy becomes one-fourth.

If kinetic energy becomes

E′=E4E' = \frac{E}{4}

then

λ′λ=EE′=EE/4=4=2\frac{\lambda'}{\lambda} = \sqrt{\frac{E}{E'}} = \sqrt{\frac{E}{E/4}} = \sqrt{4} = 2

So,

λ′=2λ\lambda' = 2\lambda

This works because de Broglie wavelength varies inversely as the square root of kinetic energy. Therefore, the correct option is B.

Common mistakes

  • Using λ∝1E\lambda \propto \dfrac{1}{E} instead of λ∝1E\lambda \propto \dfrac{1}{\sqrt{E}}. This is wrong because momentum for a non-relativistic electron is p=2mEp = \sqrt{2mE}. First relate wavelength to momentum, then substitute the energy dependence of momentum.

  • Assuming that reducing kinetic energy to one-fourth also reduces wavelength to one-fourth. This is wrong because wavelength is inversely related to the square root of energy, so lowering energy increases the wavelength. Use the inverse square-root relation carefully.

  • Confusing direct and inverse proportionality while comparing EE and λ\lambda. Since λ∝1E\lambda \propto \dfrac{1}{\sqrt{E}}, a smaller kinetic energy gives a larger wavelength. Check whether the final trend is physically consistent before choosing the option.

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