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: Establish the inverse relationship between wavelength and energy. When an electron in a hydrogen atom transitions from a higher energy level ($n_2$) to a lower one ($n_1$), it emits a photon with energy $E = hf = \frac{hc}{\lambda}$. Consequently, $\lambda = \frac{hc}{E}$. This indicates that a longer wavelength ($\lambda$) corresponds to a smaller energy difference ($E$). Therefore, the transition with the least energy difference will yield the longest wavelength. Step 2: Apply the energy level formula for a hydrogen atom. The energy of an electron at level $n$ in a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} \, \text{eV}$. The energy difference for a transition from $n_2$ to $n_1$ (where $n_2>n_1$) is calculated as $\Delta E = E_{n_2} - E_{n_1} = -\frac{13.6}{n_2^2} - \left(-\frac{13.6}{n_1^2}\right) = 13.6 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$. Step 3: Compute the energy difference for each transition. - (A) Transition from $n = 4$ 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) Transition from $n = 3$ 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) Transition from $n = 2$ 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) Transition from $n = 3$ 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 calculated 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 observed for transition (A) from $n = 4$ to $n = 3$. This transition will therefore have the largest wavelength. Step 5: Verify with the Rydberg formula (optional). The Rydberg formula for wavenumber is $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$. A smaller value of $\frac{1}{\lambda}$ implies a larger $\lambda$. The term $\frac{1}{n_1^2} - \frac{1}{n_2^2}$ is minimized for transition (A), confirming the result obtained earlier.