MCQMediumJEE 2025Bohr's Model & Hydrogen Spectrum

JEE Physics 2025 Question with Solution

Considering Bohr’s atomic model for hydrogen atom :

  • A

    (B), (C) only

  • B

    (A), (B) only

  • C

    (A), (D) only

  • D

    (A), (C) only

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: We need to identify the correct statements using Bohr's atomic model for hydrogen-like species.

Find: Which pair among (A), (B), (C), (D) is correct.

For a hydrogen-like atom, the energy of the electron in the nthn^{\text{th}} orbit is

En=13.6Z2n2eVE_n = -13.6\frac{Z^2}{n^2} \, \text{eV}

where ZZ is the atomic number.

Also,

  • ground state n=1\Rightarrow n=1
  • first excited state n=2\Rightarrow n=2
  • second excited state n=3\Rightarrow n=3

Now evaluate the relevant comparisons shown in the solution.

For H atom in ground state:

EH, ground=13.6×1212=13.6eVE_{\text{H, ground}} = -13.6 \times \frac{1^2}{1^2} = -13.6 \, \text{eV}

For He+^+ in first excited state:

EHe+,1st excited=13.6×2222=13.6eVE_{\text{He}^+, \text{1st excited}} = -13.6 \times \frac{2^2}{2^2} = -13.6 \, \text{eV}

So statement (A) is true.

For Li2+^{2+} in second excited state:

ELi2+,2nd excited=13.6×3232=13.6eVE_{\text{Li}^{2+}, \text{2nd excited}} = -13.6 \times \frac{3^2}{3^2} = -13.6 \, \text{eV}

Thus this is also equal to the ground-state energy of hydrogen, so statement (B) is true.

For He+^+ in ground state:

EHe+,ground=13.6×2212=54.4eVE_{\text{He}^+, \text{ground}} = -13.6 \times \frac{2^2}{1^2} = -54.4 \, \text{eV}

This is not equal to 13.6eV-13.6 \, \text{eV}, so statement (C) is false.

For Li2+^{2+} in ground state:

ELi2+,ground=13.6×3212=122.4eVE_{\text{Li}^{2+}, \text{ground}} = -13.6 \times \frac{3^2}{1^2} = -122.4 \, \text{eV}

This is not equal to the first excited state energy of He+^+, so statement (D) is false.

Therefore, only (A) and (B) are correct. The correct option is B.

Statement-wise Evaluation

Given: The question is based on Bohr energy levels for hydrogen-like atoms.

Find: Which statements among (A), (B), (C), (D) are correct.

The key dependence is

EnZ2n2E_n \propto \frac{Z^2}{n^2}

So whenever Z2n2\frac{Z^2}{n^2} is the same for two cases, their energies are equal.

  • For H ground state: Z2n2=1212=1\frac{Z^2}{n^2} = \frac{1^2}{1^2} = 1
  • For He+^+ first excited state: 2222=1\frac{2^2}{2^2} = 1
  • For Li2+^{2+} second excited state: 3232=1\frac{3^2}{3^2} = 1

Hence these three states have the same energy, namely 13.6eV-13.6 \, \text{eV}.

So:

  • (A) is correct
  • (B) is correct

Now check the remaining two:

  • He+^+ ground state gives 2212=4\frac{2^2}{1^2} = 4, so its energy is much lower, not equal to hydrogen ground state. Hence (C) is false.
  • Li2+^{2+} ground state gives 3212=9\frac{3^2}{1^2} = 9, so it is also not equal to He+^+ first excited state. Hence (D) is false.

Therefore, the correct option is B, that is (A), (B) only.

Common mistakes

  • Using EnZn2E_n \propto \frac{Z}{n^2} instead of the correct Bohr relation EnZ2n2E_n \propto \frac{Z^2}{n^2}. This gives wrong comparisons for hydrogen-like ions. Always use the square of the atomic number.

  • Mixing up ground state, first excited state, and second excited state. They correspond to n=1n=1, n=2n=2, and n=3n=3 respectively. Use the correct value of nn before substituting in the formula.

  • Comparing only the sign or only the magnitude loosely without calculating the actual energy. In Bohr model, all bound-state energies are negative, so equality must be checked by full numerical evaluation, not by sign alone.

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