NEET 2022 Physics Bohr Model – Energy of Electron MCQ Question

Chapter: Atoms

Class: 12 Physics Topic: Bohr Model – Energy of Electron Difficulty: 🟡 Medium

✅ Ans: B — $9:4$

For hydrogen atom, energy of electron in the $n^{th}$ orbit is:

$$E_n=-\frac{13.6}{n^2}\text{ eV}$$

• First excited state: $n=2$

$$T_1=E_2=-\frac{13.6}{4}$$

• Second excited state: $n=3$

$$T_2=E_3=-\frac{13.6}{9}$$

Therefore,

$$\frac{T_1}{T_2} = \frac{-13.6/4}{-13.6/9} =\frac94$$ $$\boxed{T_1:T_2=9:4}$$

❌ Why other options are wrong?

• A: $4:9$ ❌ This is the ratio of $1/n^2$, ignoring the fact that $T_1$ and $T_2$ are being compared as the total energies with $n=2,3$.
• B: $9:4$ ✅ Correct.
• C: $1:4$ ❌ Incorrect application of the Bohr energy relation.
• D: $4:1$ ❌ Inverse of the required ratio.

📌 NCERT Concept

For a hydrogen atom:

$$\boxed{E_n=-\frac{13.6}{n^2}\text{ eV}}$$

The negative sign indicates that the electron is bound to the nucleus.

🧠 NEET Trick

First excited → $n=2$ Second excited → $n=3$

$$E_2:E_3=\frac1{2^2}:\frac1{3^2} = \boxed{9:4}$$

Final Answer: $\boxed{B\;(9:4)}$