Question

Out of the four options given, in which transition will the emitted photon have the maximum wavelength?

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

Hint:

For hydrogen atom transitions, the photon with the maximum wavelength corresponds to the smallest energy difference. Use \( \Delta E = 13.6 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \), and the smallest \( \Delta E \) occurs for transitions between the closest energy levels.

Answer 1

The Correct Option is A

Step 1: Understand the relationship between wavelength and energy.
When an electron in a hydrogen atom transitions from a higher energy level \( n_2 \) to a lower energy level \( n_1 \), it emits a photon. The energy of the photon is: \[ E = h f = \frac{h c}{\lambda} \] \[ \lambda = \frac{h c}{E} \] A larger wavelength \( \lambda \) corresponds to a smaller energy difference \( E \). Thus, the transition with the smallest energy difference will have the maximum wavelength. Step 2: Use the energy level formula for a hydrogen atom.
The energy of an electron in a hydrogen atom at level \( n \) is: \[ E_n = -\frac{13.6}{n^2} \, \text{eV} \] The energy difference for a transition from \( n_2 \) to \( n_1 \) (\( n_2>n_1 \)) is: \[ \Delta E = E_{n_2} - E_{n_1} = -\frac{13.6}{n_2^2} - \left(-\frac{13.6}{n_1^2}\right) = 13.6 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] Step 3: Calculate the energy difference for each transition.
- (A) \( n = 4 \) to \( n = 3 \): \[ \Delta E = 13.6 \left( \frac{1}{3^2} - \frac{1}{4^2} \right) = 13.6 \left( \frac{1}{9} - \frac{1}{16} \right) = 13.6 \left( \frac{16 - 9}{144} \right) = 13.6 \times \frac{7}{144} \approx 0.661 \, \text{eV} \] - (B) \( n = 3 \) to \( n = 2 \): \[ \Delta E = 13.6 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = 13.6 \left( \frac{1}{4} - \frac{1}{9} \right) = 13.6 \left( \frac{9 - 4}{36} \right) = 13.6 \times \frac{5}{36} \approx 1.889 \, \text{eV} \] - (C) \( n = 2 \) to \( n = 1 \): \[ \Delta E = 13.6 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = 13.6 \left( \frac{1}{1} - \frac{1}{4} \right) = 13.6 \times \frac{3}{4} = 10.2 \, \text{eV} \] - (D) \( n = 3 \) to \( n = 1 \): \[ \Delta E = 13.6 \left( \frac{1}{1^2} - \frac{1}{3^2} \right) = 13.6 \left( \frac{1}{1} - \frac{1}{9} \right) = 13.6 \times \frac{8}{9} \approx 12.089 \, \text{eV} \] Step 4: Compare the energy differences.
- (A) \( \Delta E \approx 0.661 \, \text{eV} \) - (B) \( \Delta E \approx 1.889 \, \text{eV} \) - (C) \( \Delta E = 10.2 \, \text{eV} \) - (D) \( \Delta E \approx 12.089 \, \text{eV} \) The smallest energy difference is for (A) \( n = 4 \) to \( n = 3 \), which means it will have the largest wavelength. Step 5: Confirm using the Rydberg formula (optional).
The Rydberg formula for the wave number is: \[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] A smaller \( \frac{1}{\lambda} \) means a larger \( \lambda \). The term \( \frac{1}{n_1^2} - \frac{1}{n_2^2} \) is smallest for (A), confirming our result.