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MCQEasyJEE 2024Bohr's Model & Hydrogen Spectrum

JEE Physics 2024 Question with Solution

The ratio of the magnitude of the kinetic energy (KE) to the potential energy (PE) of an electron in the **55**th excited state of a hydrogen atom is:

  • A

    44

  • B

    1/41/4

  • C

    1/21/2

  • D

    11

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: An electron is in the **55**th excited state of a hydrogen atom.

Find: The ratio of the magnitude of KE to PE.

In the Bohr model of the hydrogen atom, the electrostatic attraction provides the centripetal force:

mv2r=14πϵ0e2r2\frac{mv^2}{r} = \frac{1}{4\pi\epsilon_0} \frac{e^2}{r^2}

So, the kinetic energy is

KE=12mv2=e28πϵ0rKE = \frac{1}{2}mv^2 = \frac{e^2}{8\pi\epsilon_0 r}

and the potential energy is

PE=−e24πϵ0rPE = -\frac{e^2}{4\pi\epsilon_0 r}

Taking the magnitude of potential energy,

∣PE∣=e24πϵ0r|PE| = \frac{e^2}{4\pi\epsilon_0 r}

Now form the required ratio:

KE∣PE∣=e28πϵ0re24πϵ0r=12\frac{KE}{|PE|} = \frac{\frac{e^2}{8\pi\epsilon_0 r}}{\frac{e^2}{4\pi\epsilon_0 r}} = \frac{1}{2}

This ratio is the same for every orbit, so the **55**th excited state does not change it. Therefore, the correct option is C and the ratio is 1/21/2.

Direct Relation

Given: Bohr model of hydrogen atom.

Find: KE∣PE∣\frac{KE}{|PE|}

Use the standard Bohr-model relation:

∣PE∣=2×KE|PE| = 2 \times KE

Hence,

KE∣PE∣=12\frac{KE}{|PE|} = \frac{1}{2}

Therefore, the correct option is C.

Common mistakes

  • Using KEPE\frac{KE}{PE} instead of KE∣PE∣\frac{KE}{|PE|} is incorrect because potential energy is negative in the Bohr model. First take the magnitude of potential energy, then form the ratio.

  • Assuming the answer depends on the **55**th excited state is incorrect here. The relation PE=−2KEPE = -2KE holds for every allowed orbit of hydrogen, so the ratio remains constant.

  • Confusing 5th excited state with n=5n=5 is incorrect because the **55**th excited state corresponds to n=6n=6. Although this does not affect the final ratio here, the level counting should still be done correctly.

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