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: Understanding the Concept:
The wavenumber ($\bar{\nu}$) of emitted radiation is calculated using the Rydberg formula.
Step 2: Key Formula or Approach:
$\bar{\nu} = \text{R}_\text{H} \left[\frac{1}{\text{n}_1^2} - \frac{1}{\text{n}_2^2}\right]$ where $\text{n}_1<\text{n}_2$.
Step 3: Detailed Explanation:
Given: $\text{n}_1 = 2$, $\text{n}_2 = 5$, $\text{R}_\text{H} = 109677\text{ cm}^{-1}$.
\[ \bar{\nu} = 109677 \times \left[\frac{1}{2^2} - \frac{1}{5^2}\right] \]
\[ \bar{\nu} = 109677 \times \left[\frac{1}{4} - \frac{1}{25}\right] \]
\[ \bar{\nu} = 109677 \times \left[\frac{21}{100}\right] \]
\[ \bar{\nu} = 1096.77 \times 21 = 23032.17\text{ cm}^{-1} \approx 23032\text{ cm}^{-1} \]
Step 4: Final Answer:
The wavenumber is $23032\text{ cm}^{-1}$.