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Atoms and Nuclei question

2025 · 2 Apr · Shift 1 · Q63
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Atoms and Nuclei question

2025 · 2 Apr · Shift 1 · Q63

JEE MainPhysicsAtoms and NucleiMCQ+4 / −1
Considering Bohr's atomic model for hydrogen atom : (A) the energy of H atom in ground state is same as energy of He+\mathrm{He}^{+}He+ ion in its first excited state. (B) the energy of H atom in ground state is same as that for Li++\mathrm{Li}^{++}Li++ ion in its second excited state. (C) the energy of H atom in its ground state is same as that of He+\mathrm{He}^{+}He+ ion for its ground state. (D) the energy of He+\mathrm{He}^{+}He+ ion in its first excited state is same as that for Li++\mathrm{Li}^{++}Li++ ion in its ground state. Choose the correct answer from the options given below :
  1. A
    (A), (B) only
  2. B
    (A), (D) only
  3. C
    (A), (C) only
  4. D
    (B), (D) only
View written solutionFree

Correct answer: A

  1. Use Bohr's energy formula for hydrogen-like species

For a hydrogen-like atom/ion,

En=−13.6 Z2n2 eVE_n = -13.6\,\frac{Z^2}{n^2}\ \text{eV}En​=−13.6n2Z2​ eV

where:

  • ZZZ = atomic number
  • nnn = principal quantum number

For the given species:

  • Hydrogen: Z=1Z=1Z=1
  • He+\mathrm{He}^+He+: Z=2Z=2Z=2
  • Li++\mathrm{Li}^{++}Li++: Z=3Z=3Z=3

  1. Find the required energies

(i) Hydrogen in ground state

Ground state means n=1n=1n=1.

E(H,n=1)=−13.61212=−13.6 eVE(\mathrm{H}, n=1) = -13.6\frac{1^2}{1^2} = -13.6\ \text{eV}E(H,n=1)=−13.61212​=−13.6 eV

(ii) He+\mathrm{He}^+He+ in first excited state

First excited state means n=2n=2n=2.

E(He+,n=2)=−13.62222=−13.6 eVE(\mathrm{He}^+, n=2) = -13.6\frac{2^2}{2^2} = -13.6\ \text{eV}E(He+,n=2)=−13.62222​=−13.6 eV

So statement (A) is true.

(iii) Li++\mathrm{Li}^{++}Li++ in second excited state

Second excited state means n=3n=3n=3.

E(Li++,n=3)=−13.63232=−13.6 eVE(\mathrm{Li}^{++}, n=3) = -13.6\frac{3^2}{3^2} = -13.6\ \text{eV}E(Li++,n=3)=−13.63232​=−13.6 eV

So statement (B) is true.

(iv) He+\mathrm{He}^+He+ in ground state

Ground state means n=1n=1n=1.

E(He+,n=1)=−13.62212=−54.4 eVE(\mathrm{He}^+, n=1) = -13.6\frac{2^2}{1^2} = -54.4\ \text{eV}E(He+,n=1)=−13.61222​=−54.4 eV

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

(v) Compare He+\mathrm{He}^+He+ first excited state and Li++\mathrm{Li}^{++}Li++ ground state

We already have:

E(He+,n=2)=−13.6 eVE(\mathrm{He}^+, n=2) = -13.6\,\text{eV}E(He+,n=2)=−13.6eV

For Li++\mathrm{Li}^{++}Li++ ground state, n=1n=1n=1:

E(Li++,n=1)=−13.63212=−122.4 eVE(\mathrm{Li}^{++}, n=1) = -13.6\frac{3^2}{1^2} = -122.4\ \text{eV}E(Li++,n=1)=−13.61232​=−122.4 eV

These are not equal. So statement (D) is false.


  1. Evaluate the options
  • (A) true
  • (B) true
  • (C) false
  • (D) false

Hence the correct choice is:

Option A: (A), (B) only\boxed{\text{Option A: (A), (B) only}}Option A: (A), (B) only​
  1. Comparison with stored correct answer

Stored correct answer = A

My derived answer = A

So, the derived answer agrees with the stored correct answer.

Next

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