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Step-by-Step Solution
Step 1: Understand the Given Information
The question states that the energy required to ionize a hydrogen-like ion in its ground state is 9 Rydbergs. We know that 1 Rydberg = 13.6 eV. We need to find the wavelength of the radiation emitted when an electron makes a transition from the second excited state (n = 3) to the ground state (n = 1).
Step 2: Calculate the Ionization Energy in eV
Ionization energy in eV:
$ \text{Ionization Energy} = 9 \times 13.6 \, \text{eV} = 122.4 \, \text{eV} $
Step 3: Relate Ionization Energy to Nuclear Charge (Z)
For a hydrogen-like ion, the ionization energy (to remove an electron from the ground state) is given by:
$ E_{\text{ionization}} = 13.6 \, Z^2 \, \text{eV} $
Equating this to 122.4 eV,
$ 13.6 \, Z^2 = 122.4 $
$ Z^2 = \frac{122.4}{13.6} = 9 $
Therefore, $ Z = 3. $
Step 4: Use the Rydberg Formula for the Emission Wavelength
The wavelength $ \lambda $ of the photon emitted when an electron transitions from energy level $ n_2 $ to $ n_1 $ (with $ n_2 > n_1 $) is given by:
$ \frac{1}{\lambda} = R \, Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) $
Step 5: Substitute n = 3 (Second Excited State) and n = 1 (Ground State)
Here, $ n_1 = 1 $ and $ n_2 = 3 $, and we have found $ Z = 3 $. Also, $ R $ (Rydberg constant) is approximately $ 1.09 \times 10^7 \, \text{m}^{-1} $.
$ \frac{1}{\lambda} = R \times 3^2 \left( \frac{1}{1^2} - \frac{1}{3^2} \right)
= R \times 9 \left( 1 - \frac{1}{9} \right)
= 9\,R \left( \frac{8}{9} \right)
= 8\,R. $
Step 6: Calculate the Wavelength
Since $ R = 1.09 \times 10^7 \, \text{m}^{-1} $,
$ \frac{1}{\lambda} = 8 \times 1.09 \times 10^7 = 8.72 \times 10^7 \, \text{m}^{-1}. $
Hence,
$ \lambda = \frac{1}{8.72 \times 10^7} \, \text{m} \approx 1.15 \times 10^{-8} \, \text{m}, $
which is about 11.4 nm.
Final Answer
The wavelength of the radiation emitted is approximately 11.4 nm.
