Question

A particular hydrogen-like ion emits the radiation of frequency $3 \times 10^{15}$ Hz when it makes a transition from $n = 2$ to $n = 1$. The frequency of radiation emitted in the transition from $n = 3$ to $n = 1$ is [\frac{x}{9} \times 10^{15} , \text{Hz}, , \text{when} , x = _____ .]

Answer 1

Given: - Frequency of radiation for transition $n = 2$ to $n = 1$: $\nu_{2 \to 1} = 3 \times 10^{15} \, \text{Hz}$ - Transition of interest: $n = 3$ to $n = 1$

Step 1: Energy Levels for Hydrogen-Like Ion

The energy difference between levels in a hydrogen-like atom is given by: $\Delta E \propto \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$ The frequency of emitted radiation is proportional to the energy difference: $\nu \propto \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$

Step 2: Calculating the Frequency Ratio

For the transition from $n = 2$ to $n = 1$:

$\nu_{2 \to 1} \propto \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = \left( 1 - \frac{1}{4} \right) = \frac{3}{4}$

For the transition from $n = 3$ to $n = 1$:

$\nu_{3 \to 1} \propto \left( \frac{1}{1^2} - \frac{1}{3^2} \right) = \left( 1 - \frac{1}{9} \right) = \frac{8}{9}$

Step 3: Finding the Frequency for $n = 3$ to $n = 1$ Transition

The ratio of the frequencies for the two transitions is:

$\frac{\nu_{3 \to 1}}{\nu_{2 \to 1}} = \frac{\frac{8}{9}}{\frac{3}{4}} = \frac{8}{9} \times \frac{4}{3} = \frac{32}{27}$

Thus:

$\nu_{3 \to 1} = \nu_{2 \to 1} \times \frac{32}{27}$

Substituting the given value:

$\nu_{3 \to 1} = 3 \times 10^{15} \times \frac{32}{27} = \frac{32}{9} \times 10^{15} \, \text{Hz}$

Step 4: Comparing with Given Expression

The frequency is given as $\frac{x}{9} \times 10^{15} \, \text{Hz}$. By comparison:

$x = 32$

Conclusion: The value of $x$ is 32.