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 = _____ .]
Correct Answer: 32
Approach Solution - 1
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 \(x = 32\) .
Approach Solution -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.
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