Find the wavelenght of the first line of `He^(+)`ion sepectral series whose interval between extreme line is
`[(1)/(lambda)-(1)/(lambda)=2.745xx10^(4)cm^(1)]`

Explanation

To find the wavelength of the first line of the `He^(+)` ion spectral series, we can use the formula for the wavelength of spectral lines in hydrogen-like atoms, which is given by the Rydberg formula:

\[ \frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

Where:
- \( R \) is the Rydberg constant (\( R = 1.097 \times 10^7 \, \text{m}^{-1} \))
- \( Z \) is the atomic number (for He, \( Z = 2 \))
- \( n_1 \) is the lower energy level
- \( n_2 \) is the higher energy level

### Step 1: Identify the values of \( n_1 \) and \( n_2 \)

For the first line of the spectral series, we typically take \( n_1 = 1 \) (ground state) and \( n_2 = 2 \) (first excited state).

### Step 2: Substitute the values into the Rydberg formula

Now, substituting \( R \), \( Z \), \( n_1 \), and \( n_2 \) into the Rydberg formula:

\[ \frac{1}{\lambda} = RZ^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) \]

Calculating the right-hand side:

\[ \frac{1}{\lambda} = (1.097 \times 10^7) \cdot (2^2) \left( 1 - \frac{1}{4} \right) \]

\[ = (1.097 \times 10^7) \cdot 4 \left( \frac{3}{4} \right) \]

\[ = (1.097 \times 10^7) \cdot 3 \]

\[ = 3.291 \times 10^7 \, \text{m}^{-1} \]

### Step 3: Calculate \( \lambda \)

Now, we can find \( \lambda \) by taking the reciprocal:

\[ \lambda = \frac{1}{3.291 \times 10^7} \, \text{m} \]

Calculating \( \lambda \):

\[ \lambda \approx 3.04 \times 10^{-8} \, \text{m} \]

To convert this to centimeters:

\[ \lambda \approx 3.04 \times 10^{-8} \, \text{m} \times 100 \, \text{cm/m} = 3.04 \times 10^{-6} \, \text{cm} \]

### Step 4: Final Result

Thus, the wavelength of the first line of the `He^(+)` ion spectral series is approximately:

\[ \lambda \approx 3.04 \times 10^{-6} \, \text{cm} \]