Calculate the Rydberg constant ` R_H` if `He^+` ions are known to have the wavelength difference between the from ( of the longest wavength ) lines fo Balmer and Lyman series equal to ` 133.7 nm`.

Explanation

To calculate the Rydberg constant \( R_H \) for the \( He^+ \) ion given the wavelength difference between the longest wavelength lines of the Balmer and Lyman series is \( 133.7 \, \text{nm} \), we can follow these steps:

### Step 1: Understand the Wavelength Formula
The formula for the wavelength in the Rydberg formula is given by:
\[ \frac{1}{\lambda} = R \cdot Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where:
- \( \lambda \) is the wavelength,
- \( R \) is the Rydberg constant,
- \( Z \) is the atomic number (for \( He^+ \), \( Z = 2 \)),
- \( n_1 \) and \( n_2 \) are the principal quantum numbers.

### Step 2: Calculate the Wavelength for the Balmer Series
For the first line of the Balmer series, the transition is from \( n_2 = 3 \) to \( n_1 = 2 \):
\[ \frac{1}{\lambda_1} = R \cdot Z^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]
Substituting \( Z = 2 \):
\[ \frac{1}{\lambda_1} = R \cdot 2^2 \left( \frac{1}{4} - \frac{1}{9} \right) \]
Calculating \( \frac{1}{4} - \frac{1}{9} \):
\[ \frac{1}{4} - \frac{1}{9} = \frac{9 - 4}{36} = \frac{5}{36} \]
Thus,
\[ \frac{1}{\lambda_1} = R \cdot 4 \cdot \frac{5}{36} = \frac{20R}{36} = \frac{5R}{9} \]
So,
\[ \lambda_1 = \frac{9}{5R} \]

### Step 3: Calculate the Wavelength for the Lyman Series
For the first line of the Lyman series, the transition is from \( n_2 = 2 \) to \( n_1 = 1 \):
\[ \frac{1}{\lambda_2} = R \cdot Z^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) \]
Substituting \( Z = 2 \):
\[ \frac{1}{\lambda_2} = R \cdot 4 \left( 1 - \frac{1}{4} \right) = R \cdot 4 \cdot \frac{3}{4} = 3R \]
Thus,
\[ \lambda_2 = \frac{1}{3R} \]

### Step 4: Calculate the Wavelength Difference
The difference in wavelengths \( \Delta \lambda \) is given by:
\[ \Delta \lambda = \lambda_1 - \lambda_2 \]
Substituting the values we found:
\[ \Delta \lambda = \frac{9}{5R} - \frac{1}{3R} \]
Finding a common denominator (15R):
\[ \Delta \lambda = \frac{27}{15R} - \frac{5}{15R} = \frac{22}{15R} \]

### Step 5: Set Up the Equation for \( R \)
We know from the problem statement that \( \Delta \lambda = 133.7 \, \text{nm} = 133.7 \times 10^{-9} \, \text{m} \):
\[ \frac{22}{15R} = 133.7 \times 10^{-9} \]
Rearranging to solve for \( R \):
\[ R = \frac{22 \times 10^{-9}}{15 \cdot 133.7} \]

### Step 6: Calculate \( R \)
Calculating \( R \):
\[ R = \frac{22 \times 10^{-9}}{2005.5} \approx 1.095 \times 10^{7} \, \text{m}^{-1} \]

### Final Answer
Thus, the Rydberg constant \( R_H \) for \( He^+ \) is approximately:
\[ R_H \approx 1.095 \times 10^{7} \, \text{m}^{-1} \]