To find the shortest wavelength of transition in the Paschen series of the `He^(+)` ion, we will use the Rydberg formula. Here’s a step-by-step solution:
### Step 1: Understand the Paschen Series
The Paschen series corresponds to transitions where the final energy level (n1) is 3, and the initial energy level (n2) is greater than 3 (n2 = 4, 5, 6, ...). The shortest wavelength corresponds to the transition from n2 = ∞ to n1 = 3.
### Step 2: Rydberg Formula
The Rydberg formula for the wavelength of emitted light during a transition is given by:
\[
\frac{1}{\lambda} = R_H \cdot Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)
\]
Where:
- \( \lambda \) is the wavelength,
- \( R_H \) is the Rydberg constant,
- \( Z \) is the atomic number,
- \( n_1 \) is the principal quantum number of the lower energy level,
- \( n_2 \) is the principal quantum number of the higher energy level.
### Step 3: Assign Values
For the `He^(+)` ion:
- The atomic number \( Z = 2 \) (since Helium has 2 protons).
- For the Paschen series, \( n_1 = 3 \) and \( n_2 = \infty \).
### Step 4: Substitute Values into the Rydberg Formula
Substituting the values into the Rydberg formula:
\[
\frac{1}{\lambda} = R_H \cdot 2^2 \left( \frac{1}{3^2} - \frac{1}{\infty^2} \right)
\]
Since \( \frac{1}{\infty^2} = 0 \), this simplifies to:
\[
\frac{1}{\lambda} = R_H \cdot 4 \cdot \frac{1}{9}
\]
\[
\frac{1}{\lambda} = \frac{4 R_H}{9}
\]
### Step 5: Use the Given Rydberg Constant
We know that \( \frac{1}{R_H} = 91.12 \, \text{nm} \), thus:
\[
R_H = \frac{1}{91.12 \, \text{nm}} \approx 0.01096 \, \text{nm}^{-1}
\]
### Step 6: Calculate \( \frac{1}{\lambda} \)
Substituting \( R_H \) into the equation:
\[
\frac{1}{\lambda} = \frac{4 \cdot 0.01096 \, \text{nm}^{-1}}{9}
\]
\[
\frac{1}{\lambda} \approx \frac{0.04384 \, \text{nm}^{-1}}{9} \approx 0.00487 \, \text{nm}^{-1}
\]
### Step 7: Calculate \( \lambda \)
Now, take the reciprocal to find \( \lambda \):
\[
\lambda \approx \frac{1}{0.00487} \approx 205.4 \, \text{nm}
\]
### Final Answer
The shortest wavelength of transition in the Paschen series of the `He^(+)` ion is approximately **205.4 nm**.
