Question

Which of the following electronic transitions in hydrogen atom will require the highest energy?

• A. \( n = 4 \) to \( n = 5 \)
• B. \( n = 1 \) to \( n = 2 \)
• C. \( n = 3 \) to \( n = 5 \)
• D. \( n = 2 \) to \( n = 3 \)

Hint:

The energy difference between adjacent energy levels in a hydrogen atom decreases as the principal quantum number \( n \) increases. Transitions between lower energy levels involve larger energy differences. The transition from the ground state (\( n=1 \)) to the first excited state (\( n=2 \)) requires the highest energy among the given options.

Answer 1

The Correct Option is B

The energy of an electron in the \( n^{th} \) orbit of a hydrogen atom is given by the formula: $$ E_n = -\frac{13.
6}{n^2} \text{ eV} $$ where \( n \) is the principal quantum number (\( n = 1, 2, 3, \ldots \)).
The energy required for an electronic transition from an initial level \( n_i \) to a final level \( n_f \) is given by the difference in their energies: $$ \Delta E = E_f - E_i = -13.
6 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \text{ eV} $$ A positive value of \( \Delta E \) indicates that energy is absorbed (required) for the transition to occur (excitation to a higher energy level, \( n_f>n_i \)).
We need to find the transition with the largest positive \( \Delta E \).
Let's calculate \( \Delta E \) for each option:
Option (A) \( n = 4 \) to \( n = 5 \): $$ \Delta E_A = -13.
6 \left( \frac{1}{5^2} - \frac{1}{4^2} \right) = -13.
6 \left( \frac{1}{25} - \frac{1}{16} \right) = -13.
6 \left( \frac{16 - 25}{400} \right) = -13.
6 \left( \frac{-9}{400} \right) = \frac{122.
4}{400} = 0.
306 \text{ eV} $$
Option (B) \( n = 1 \) to \( n = 2 \): $$ \Delta E_B = -13.
6 \left( \frac{1}{2^2} - \frac{1}{1^2} \right) = -13.
6 \left( \frac{1}{4} - 1 \right) = -13.
6 \left( -\frac{3}{4} \right) = \frac{40.
8}{4} = 10.
2 \text{ eV} $$
Option (C) \( n = 3 \) to \( n = 5 \): $$ \Delta E_C = -13.
6 \left( \frac{1}{5^2} - \frac{1}{3^2} \right) = -13.
6 \left( \frac{1}{25} - \frac{1}{9} \right) = -13.
6 \left( \frac{9 - 25}{225} \right) = -13.
6 \left( \frac{-16}{225} \right) = \frac{217.
6}{225} = 0.
967 \text{ eV} $$
Option (D) \( n = 2 \) to \( n = 3 \): $$ \Delta E_D = -13.
6 \left( \frac{1}{3^2} - \frac{1}{2^2} \right) = -13.
6 \left( \frac{1}{9} - \frac{1}{4} \right) = -13.
6 \left( \frac{4 - 9}{36} \right) = -13.
6 \left( -\frac{5}{36} \right) = \frac{68}{36} = 1.
889 \text{ eV} $$ Comparing the energy required for each transition: \( \Delta E_A = 0.
306 \) eV \( \Delta E_B = 10.
2 \) eV \( \Delta E_C = 0.
967 \) eV \( \Delta E_D = 1.
889 \) eV The highest energy required is for the transition from \( n = 1 \) to \( n = 2 \).