Question

A particular hydrogen like ion emits radiation of frequency $2.92 \times 10^{15}$ Hz when it makes transition from n=3 to n=1. The frequency in Hz of radiation emitted in transition from n=2 to n=1 will be :

• A. $4.38 \times 10^{15}$
• B. $6.57 \times 10^{15}$
• C. $0.44 \times 10^{15}$
• D. $2.46 \times 10^{15}$

Hint:

For frequency ratios, you don't need the value of $Z$ or Rydberg's constant. Simply use the term $(1/n_1^2 - 1/n_2^2)$.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The frequency of radiation emitted during an electronic transition in a hydrogen-like atom is proportional to the difference of the inverse squares of the principal quantum numbers of the levels.
Step 2: Key Formula or Approach:
\[ \nu = R c Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
For a given ion (\(Z\) is constant), \(\nu \propto \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)\).
Step 3: Detailed Explanation:
1. For transition \(n=3\) to \(n=1\):
\[ \nu_1 = k \left( \frac{1}{1^2} - \frac{1}{3^2} \right) = k \left( 1 - \frac{1}{9} \right) = \frac{8}{9} k \]
Given \(\nu_1 = 2.92 \times 10^{15} \text{ Hz}\).
2. For transition \(n=2\) to \(n=1\):
\[ \nu_2 = k \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = k \left( 1 - \frac{1}{4} \right) = \frac{3}{4} k \]
3. Finding the ratio:
\[ \frac{\nu_2}{\nu_1} = \frac{3/4}{8/9} = \frac{3}{4} \times \frac{9}{8} = \frac{27}{32} \]
\[ \nu_2 = \nu_1 \times \frac{27}{32} = (2.92 \times 10^{15}) \times 0.84375 \approx 2.46 \times 10^{15} \text{ Hz} \]
Step 4: Final Answer:
The frequency for the \(n=2\) to \(n=1\) transition is $2.46 \times 10^{15}$ Hz.