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Physics · Atomic Physics

JEE Main 2025 — 7 April, Morning Shift — Question 55

In a hydrogen like ion, the energy difference between the 2nd 2^{\text {nd }} excitation energy state and

ground is 108.8 eV . The atomic number of the ion is:

  1. Option A:

    2

  2. Option B:

    1

  3. Option C:

    3

    Correct
  4. Option D:

    4

Answer: C

Step-by-step solution

ΔE=E0(Z)2[1−19]=108.8eV\Delta E=E_{0}(Z)^{2}\left[1-\frac{1}{9}\right]=108.8 \mathrm{eV}

⇒13.6(Z)2×89=108.8⇒Z2=108.8×98×13.6=9⇒Z=3\begin{aligned} & \Rightarrow \quad 13.6(Z)^{2} \times \frac{8}{9}=108.8 & \Rightarrow \quad Z^{2}=\frac{108.8 \times 9}{8 \times 13.6}=9 & \Rightarrow \quad Z=3 \end{aligned}

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
7 April, Morning Shift
Subject
Physics
Chapter
Atomic Physics
Topic
Rutherford's and Bohr's Model of Atom

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