The atomic structure of `He^(+)` arises due to transition from `n_(2)` to `n_(1)` level. If `n_(1)+n_(2)` is 3 and `n_(2)-n_(1)` is 1. Find the `lambda` in nm of transition for this series in `He^(+)` in nm.

Answer

To solve the problem of finding the wavelength (λ) of the transition from the n=2 to n=1 energy level in the `He^(+)` ion, we will follow these steps:

### Step 1: Determine the values of n1 and n2
We are given two equations:
1. \( n_1 + n_2 = 3 \)
2. \( n_2 - n_1 = 1 \)

We can solve these equations simultaneously.

Adding the two equations:
\[ (n_1 + n_2) + (n_2 - n_1) = 3 + 1 \]
This simplifies to:
\[ 2n_2 = 4 \]
So,
\[ n_2 = \frac{4}{2} = 2 \]

Now, substituting \( n_2 = 2 \) back into the first equation:
\[ n_1 + 2 = 3 \]
Thus,
\[ n_1 = 3 - 2 = 1 \]

### Step 2: Use the Rydberg formula to find λ
The Rydberg formula for the wavelength of the emitted photon during a transition is given by:
\[ \frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
Where:
- \( R \) is the Rydberg constant, \( R = 1.1 \times 10^7 \, \text{m}^{-1} \)
- \( Z \) is the atomic number of helium, \( Z = 2 \)
- \( n_1 = 1 \)
- \( n_2 = 2 \)

Substituting the values into the Rydberg formula:
\[ \frac{1}{\lambda} = 1.1 \times 10^7 \times 2^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) \]

Calculating \( 2^2 \):
\[ 2^2 = 4 \]

Now substituting:
\[ \frac{1}{\lambda} = 1.1 \times 10^7 \times 4 \left( 1 - \frac{1}{4} \right) \]
\[ = 1.1 \times 10^7 \times 4 \left( \frac{3}{4} \right) \]

The \( 4 \) cancels out:
\[ = 1.1 \times 10^7 \times 3 \]
\[ = 3.3 \times 10^7 \, \text{m}^{-1} \]

### Step 3: Calculate λ
To find λ, we take the reciprocal:
\[ \lambda = \frac{1}{3.3 \times 10^7} \]

Calculating this gives:
\[ \lambda \approx 0.303 \times 10^{-7} \, \text{m} \]

### Step 4: Convert λ to nanometers
To convert meters to nanometers:
\[ 1 \, \text{m} = 10^9 \, \text{nm} \]
Thus,
\[ \lambda \approx 0.303 \times 10^{-7} \times 10^9 \, \text{nm} \]
\[ = 30.3 \, \text{nm} \]

### Final Answer
The wavelength (λ) of the transition for this series in `He^(+)` is approximately **30 nm**.