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

The Rydberg formula is employed to determine the frequency of emitted radiation during electron transitions in hydrogen-like ions: \[ u = RZ^2\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) \] Here, \( u \) represents frequency, \( R \) is the Rydberg constant (\(1.097 \times 10^7 \, \text{m}^{-1}\)), \( Z \) denotes the atomic number, and \( n_1 \) and \( n_2 \) are the principal quantum numbers of the involved energy levels.

Given \( u_{2 \to 1} = 3 \times 10^{15} \, \text{Hz} \), the following is obtained:

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

Using this relationship, we calculate \( u_{3 \to 1} \):

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

From the given \( u_{2 \to 1} = 3 \times 10^{15} \, \text{Hz} \), we have:

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

Solving for \( RZ^2 \), we get:

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

Substituting this value into the equation for \( u_{3 \to 1} \):

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

To match the requested format \(\frac{x}{9} \times 10^{15}\text{ Hz}\), \(x = 32\). Therefore, \(x = 32\).