A hydrogen-like atom emits radiation of frequency `2.7 xx 10^(15)` Hz when if makes a transition from n = 2 to n = 1. The frequency emitted in a transition from n = 3 to n = 1 will be

Options

A. `1.8xx10^(15)` Hz
B. `3.2xx10^(15)` Hz
C. `4.7xx10^(15)` Hz
D. `6.9xx10^(15)` Hz

Explanation

To solve the problem, we need to find the frequency emitted during a transition from n = 3 to n = 1 in a hydrogen-like atom, given that the frequency emitted during a transition from n = 2 to n = 1 is \(2.7 \times 10^{15}\) Hz.

### Step-by-Step Solution:

1. **Understanding the Formula for Frequency Emission:**
The frequency of radiation emitted during a transition between two energy levels in a hydrogen-like atom can be expressed as:
\[ \nu = k \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where:
- \(k\) is a constant that depends on the atomic number \(Z\) and the speed of light \(c\).
- \(n_1\) is the principal quantum number of the final state.
- \(n_2\) is the principal quantum number of the initial state.

2. **Calculating the First Transition (n = 2 to n = 1):**
For the transition from \(n = 2\) to \(n = 1\):
- \(n_1 = 1\)
- \(n_2 = 2\)

Substituting these values into the frequency formula:
\[ \nu_1 = k \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = k \left( 1 - \frac{1}{4} \right) = k \left( \frac{3}{4} \right) \]
Given that \(\nu_1 = 2.7 \times 10^{15}\) Hz, we have:
\[ 2.7 \times 10^{15} = k \left( \frac{3}{4} \right) \]

3. **Calculating the Second Transition (n = 3 to n = 1):**
For the transition from \(n = 3\) to \(n = 1\):
- \(n_1 = 1\)
- \(n_2 = 3\)

Substituting these values into the frequency formula:
\[ \nu_2 = k \left( \frac{1}{1^2} - \frac{1}{3^2} \right) = k \left( 1 - \frac{1}{9} \right) = k \left( \frac{8}{9} \right) \]

4. **Finding the Ratio of Frequencies:**
To find the frequency \(\nu_2\) in terms of \(\nu_1\), we can set up the ratio:
\[ \frac{\nu_2}{\nu_1} = \frac{k \left( \frac{8}{9} \right)}{k \left( \frac{3}{4} \right)} = \frac{8/9}{3/4} = \frac{8 \cdot 4}{9 \cdot 3} = \frac{32}{27} \]

5. **Calculating \(\nu_2\):**
Now we can express \(\nu_2\) in terms of \(\nu_1\):
\[ \nu_2 = \nu_1 \cdot \frac{32}{27} = \left( 2.7 \times 10^{15} \right) \cdot \frac{32}{27} \]

Performing the calculation:
\[ \nu_2 = 2.7 \times 10^{15} \cdot \frac{32}{27} \approx 3.2 \times 10^{15} \text{ Hz} \]

### Final Answer:
The frequency emitted during the transition from \(n = 3\) to \(n = 1\) is approximately:
\[ \boxed{3.2 \times 10^{15} \text{ Hz}} \]