The wavelength of `H_(alpha)` line of Balmer series is `X Å` what is the `X of H_(beta)`line of Balmer series

Options

A. `X (108)/(80) Å`
B. `X (80)/(108) Å`
C. `(1)/(X) (80)/(108) Å`
D. `(1)/(X) (108)/(80) Å`

Explanation

To find the wavelength of the H_beta line of the Balmer series given that the wavelength of the H_alpha line is X Å, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Balmer Series Formula**:
The formula for the wavelength (λ) of the Balmer series is given by:
\[ \frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{n^2} \right) \]
where \( R \) is the Rydberg constant and \( n \) is the principal quantum number for the higher energy level.

2. **Finding the Wavelength for H_alpha**:
For the H_alpha line, the transition is from \( n = 3 \) to \( n = 2 \).
\[ \frac{1}{x} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]
Simplifying this:
\[ \frac{1}{x} = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{9 - 4}{36} \right) = \frac{5R}{36} \]
Rearranging gives:
\[ R = \frac{36}{5x} \]

3. **Finding the Wavelength for H_beta**:
For the H_beta line, the transition is from \( n = 4 \) to \( n = 2 \).
\[ \frac{1}{\lambda_{H\beta}} = R \left( \frac{1}{2^2} - \frac{1}{4^2} \right) \]
Simplifying this:
\[ \frac{1}{\lambda_{H\beta}} = R \left( \frac{1}{4} - \frac{1}{16} \right) = R \left( \frac{4 - 1}{16} \right) = \frac{3R}{16} \]

4. **Substituting R into the H_beta Equation**:
Now substitute the value of \( R \) from the H_alpha equation into the H_beta equation:
\[ \frac{1}{\lambda_{H\beta}} = \frac{3}{16} \left( \frac{36}{5x} \right) \]
Simplifying this:
\[ \frac{1}{\lambda_{H\beta}} = \frac{108}{80x} \]
Therefore:
\[ \lambda_{H\beta} = \frac{80x}{108} \]

5. **Final Result**:
Thus, the wavelength of the H_beta line is:
\[ \lambda_{H\beta} = \frac{80}{108} x \text{ Å} \]