H-like atom with ionization energy of 9R. Find the wavelength of light emitted (in nm) when electron jumps from second excited state to ground state. (R is Rydberg constant)

Options

A. `35.8 nm`
B. `24.2 nm`
C. `8.6 nm`
D. `11.4nm `

Answer

To solve the problem, we need to find the wavelength of light emitted when an electron in a hydrogen-like atom jumps from the second excited state (n=3) to the ground state (n=1). The ionization energy is given as 9R, where R is the Rydberg constant.

### Step-by-step Solution:

1. **Identify the Ionization Energy Formula:**
The ionization energy (IE) for a hydrogen-like atom can be expressed as:
\[ IE = \frac{RZ^2}{n^2} \]
where \( Z \) is the atomic number and \( n \) is the principal quantum number.

2. **Set Up the Equation:**
Given that the ionization energy is \( 9R \), we can set up the equation:
\[ 9R = \frac{RZ^2}{1^2} \]
Here, we consider the ground state (n=1) for the ionization energy.

3. **Solve for Z:**
By canceling \( R \) from both sides, we have:
\[ 9 = Z^2 \]
Taking the square root gives:
\[ Z = 3 \]

4. **Calculate the Energy Difference (ΔE):**
The energy difference when the electron jumps from n=3 to n=1 is given by:
\[ \Delta E = 13.6 \times Z^2 \left( \frac{1}{1^2} - \frac{1}{3^2} \right) \]
Substituting \( Z = 3 \):
\[ \Delta E = 13.6 \times 3^2 \left( 1 - \frac{1}{9} \right) \]
Simplifying further:
\[ \Delta E = 13.6 \times 9 \left( 1 - \frac{1}{9} \right) = 13.6 \times 9 \times \frac{8}{9} \]
The \( 9 \) cancels out:
\[ \Delta E = 13.6 \times 8 = 108.8 \text{ eV} \]

5. **Calculate the Wavelength (λ):**
Using the relationship between energy and wavelength:
\[ E = \frac{hc}{\lambda} \]
Rearranging gives:
\[ \lambda = \frac{hc}{\Delta E} \]
Substituting \( h = 6.626 \times 10^{-34} \text{ J s} \) and \( c = 3 \times 10^8 \text{ m/s} \), and converting \( \Delta E \) to Joules (1 eV = \( 1.6 \times 10^{-19} \) J):
\[ \Delta E = 108.8 \text{ eV} \times 1.6 \times 10^{-19} \text{ J/eV} = 1.7408 \times 10^{-17} \text{ J} \]
Now substituting the values:
\[ \lambda = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{1.7408 \times 10^{-17}} \]
Calculating gives:
\[ \lambda \approx 1.14 \times 10^{-8} \text{ m} = 11.4 \text{ nm} \]

### Final Answer:
The wavelength of light emitted when the electron jumps from the second excited state to the ground state is approximately **11.4 nm**.