Q185 / 200 · MHT CET - Medical 2026 (25 April Shift 1) · Physics
The ratio of ionization energy of a hydrogen atom in fifth excited state to second excited state is
- A$1:1$
- B$1:2$
- C$4:1$
- Correct answerD$1:4$
Solution
The ionization energy of a hydrogen atom in the $n$-th state is given by $IE_n = \dfrac{13.6}{n^2}$ eV.
For the fifth excited state, the principal quantum number is $n_1 = 6$.
For the second excited state, the principal quantum number is $n_2 = 3$.
The ratio of their ionization energies is:
$\dfrac{IE_6}{IE_3} = \dfrac{\dfrac{13.6}{6^2}}{\dfrac{13.6}{3^2}}$
$\dfrac{IE_6}{IE_3} = \dfrac{3^2}{6^2} = \dfrac{9}{36} = \dfrac{1}{4}$
The ratio is $1:4$.
Answer: $1:4$
For the fifth excited state, the principal quantum number is $n_1 = 6$.
For the second excited state, the principal quantum number is $n_2 = 3$.
The ratio of their ionization energies is:
$\dfrac{IE_6}{IE_3} = \dfrac{\dfrac{13.6}{6^2}}{\dfrac{13.6}{3^2}}$
$\dfrac{IE_6}{IE_3} = \dfrac{3^2}{6^2} = \dfrac{9}{36} = \dfrac{1}{4}$
The ratio is $1:4$.
Answer: $1:4$
Hence, the correct answer is option: 'D'.