Question

What is the wavenumber of the photon emitted during transition from the orbit $\text{n} = 5$ to that of $\text{n} = 2$ in hydrogen atom? $[ \text{R}_\text{H} = 109677\text{ cm}^{-1} ]$}

• A. $23032\text{ cm}^{-1}$
• B. $46064\text{ cm}^{-1}$
• C. $69096\text{ cm}^{-1}$
• D. $92128\text{ cm}^{-1}$

Hint:

Transition to $n=2$ corresponds to the Balmer series.

Answer 1

The Correct Option is A

Step 1: Concept Use the Rydberg formula for wavenumber ($\bar{\nu}$): $\bar{\nu} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.

Step 2: Meaning Here, $n_1 = 2$ (lower level) and $n_2 = 5$ (higher level).

Step 3: Analysis $\bar{\nu} = 109677 \left( \frac{1}{2^2} - \frac{1}{5^2} \right) = 109677 \left( \frac{1}{4} - \frac{1}{25} \right)$.
$\bar{\nu} = 109677 \left( \frac{21}{100} \right) = 109677 \times 0.21$.

Step 4: Conclusion $\bar{\nu} \approx 23032.17\text{ cm}^{-1}$. Final Answer: (A)