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 can follow these steps:

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

Let's solve these equations to find the values of \( n_1 \) and \( n_2 \).

### Step 2: Solve the equations
From the second equation, we can express \( n_2 \) in terms of \( n_1 \):
\[ n_2 = n_1 + 1 \]

Now, substitute this expression into the first equation:
\[ n_1 + (n_1 + 1) = 3 \]
\[ 2n_1 + 1 = 3 \]
\[ 2n_1 = 2 \]
\[ n_1 = 1 \]

Now substitute \( n_1 \) back to find \( n_2 \):
\[ n_2 = n_1 + 1 = 1 + 1 = 2 \]

Thus, we have:
- \( n_1 = 1 \)
- \( n_2 = 2 \)

### Step 3: Use Rydberg's formula
For hydrogen-like ions, the Rydberg formula 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 (for He, \( Z = 2 \))
- \( n_1 = 1 \) and \( n_2 = 2 \)

### Step 4: Substitute the values into the formula
Substituting the known values into the Rydberg formula:
\[ \frac{1}{\lambda} = 1.1 \times 10^7 \times 2^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) \]
\[ = 1.1 \times 10^7 \times 4 \left( 1 - \frac{1}{4} \right) \]
\[ = 1.1 \times 10^7 \times 4 \left( \frac{3}{4} \right) \]

### Step 5: Simplify the expression
\[ \frac{1}{\lambda} = 1.1 \times 10^7 \times 3 \]
\[ = 3.3 \times 10^7 \, \text{m}^{-1} \]

### Step 6: Calculate λ
Taking the reciprocal to find λ:
\[ \lambda = \frac{1}{3.3 \times 10^7} \]
\[ \lambda \approx 3.03 \times 10^{-8} \, \text{m} \]

### Step 7: Convert λ to nanometers
To convert meters to nanometers, we multiply by \( 10^9 \):
\[ \lambda \approx 3.03 \times 10^{-8} \, \text{m} \times 10^9 \]
\[ \lambda \approx 30.3 \, \text{nm} \]

### Final Answer
The wavelength of the transition from n=2 to n=1 in He⁺ is approximately **30.3 nm**.

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