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Step-by-Step Solution
Step 1: Recall the frequency relation for hydrogen-like ions
The frequency of radiation emitted during a transition in a hydrogen-like ion is proportional to the difference in the inverse squares of the principal quantum numbers. Mathematically, for a transition from an initial level $n_i$ to a final level $n_f$, the frequency $f$ is given by:
$f \propto \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$
So if we have two transitions, their frequencies can be compared using the ratio of these differences.
Step 2: Write expressions for given transitions
Let $f_1$ be the frequency for the transition from $n = 3$ to $n = 1$, and $f_2$ for the transition from $n = 2$ to $n = 1$. Then:
$f_1 \propto \left(\frac{1}{1^2} - \frac{1}{3^2}\right) = 1 - \frac{1}{9} = \frac{8}{9}
$
$f_2 \propto \left(\frac{1}{1^2} - \frac{1}{2^2}\right) = 1 - \frac{1}{4} = \frac{3}{4}
$
Step 3: Form the ratio of the two frequencies
Since $f \propto \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)$, the ratio of the two frequencies is:
$\frac{f_1}{f_2} = \frac{\frac{8}{9}}{\frac{3}{4}} = \frac{8}{9} \times \frac{4}{3} = \frac{32}{27}
$
Step 4: Substitute the known frequency and solve for $f_2$
We are given $f_1 = 2.92 \times 10^{15}\,\text{Hz}$. Therefore,
$f_2 = f_1 \times \frac{f_2}{f_1} = 2.92 \times 10^{15} \times \left(\frac{3/4}{8/9}\right) = 2.92 \times 10^{15} \times \frac{3 \times 9}{4 \times 8} = 2.92 \times 10^{15} \times \frac{27}{32}.
Numerically,
$f_2 \approx 2.46 \times 10^{15}\,\text{Hz}.
Step 5: Final Answer
The frequency of the radiation emitted in the transition from $n = 2$ to $n = 1$ is
$\boxed{2.46 \times 10^{15}\,\text{Hz}.}
