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

Correct Answer: 32

To solve this problem, we use the Rydberg formula for hydrogen-like ions to calculate the frequency of emitted radiation during electron transitions: \[ \nu = RZ^2\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) \] where \( \nu \) is the frequency, \( R \) is the Rydberg constant (\(1.097 \times 10^7 \, \text{m}^{-1}\)), \( Z \) is the atomic number, and \( n_1 \) and \( n_2 \) are the principal quantum numbers of the two levels.

Given \( \nu_{2 \to 1} = 3 \times 10^{15} \, \text{Hz} \), we find: 

\[ \nu_{2 \to 1} = RZ^2\left(\frac{1}{1^2} - \frac{1}{2^2}\right) = RZ^2\left(\frac{3}{4}\right) \]

Using this as a basis, let's calculate \( \nu_{3 \to 1} \):

\[ \nu_{3 \to 1} = RZ^2\left(\frac{1}{1^2} - \frac{1}{3^2}\right) = RZ^2\left(\frac{8}{9}\right) \]

From \( \nu_{2 \to 1} = 3 \times 10^{15} \, \text{Hz} \),

\[ 3 \times 10^{15} = RZ^2\left(\frac{3}{4}\right) \]

Solving for \( RZ^2 \),

\[ RZ^2 = \frac{4 \times 3 \times 10^{15}}{3} = 4 \times 10^{15} \]

Substitute this into \( \nu_{3 \to 1} \):

\[ \nu_{3 \to 1} = 4 \times 10^{15}\left(\frac{8}{9}\right) = \frac{32}{9} \times 10^{15} \, \text{Hz} \]

Comparing this to the requested format \(\frac{x}{9} \times 10^{15}\text{ Hz}\), \(x = 32\). The value \(x = 32\) .

Answer 2

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.