Wavelength of the Balmer H, line (first line) is `6565 Å`. Calculate the wavelength of `H_(beta)` (second line).

Explanation

To calculate the wavelength of the H_beta line (the second line) in the Balmer series of hydrogen, we will follow these steps:

### Step 1: Understand the Balmer Series
The Balmer series corresponds to electronic transitions in hydrogen where the electron falls to the n=2 energy level from higher energy levels (n=3, 4, 5, ...). The first line (H_alpha) corresponds to the transition from n=3 to n=2, and the second line (H_beta) corresponds to the transition from n=4 to n=2.

### Step 2: Use the Rydberg Formula
The Rydberg formula for the wavelength of spectral lines in hydrogen is given by:
\[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where:
- \( R_H \) is the Rydberg constant (approximately \( 1.097 \times 10^7 \, \text{m}^{-1} \)),
- \( n_1 \) is the lower energy level,
- \( n_2 \) is the higher energy level.

### Step 3: Calculate for H_alpha (First Line)
For H_alpha (first line):
- \( n_1 = 2 \)
- \( n_2 = 3 \)

Using the Rydberg formula:
\[ \frac{1}{\lambda_{H_\alpha}} = R_H \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]
Substituting the values:
\[ \frac{1}{\lambda_{H_\alpha}} = R_H \left( \frac{1}{4} - \frac{1}{9} \right) \]
Calculating the right side:
\[ \frac{1}{4} - \frac{1}{9} = \frac{9 - 4}{36} = \frac{5}{36} \]
Thus,
\[ \frac{1}{\lambda_{H_\alpha}} = R_H \cdot \frac{5}{36} \]

### Step 4: Substitute the Known Wavelength for H_alpha
Given that the wavelength of H_alpha is \( 6565 \, \text{Å} \):
\[ \frac{1}{6565} = R_H \cdot \frac{5}{36} \]

### Step 5: Calculate for H_beta (Second Line)
For H_beta (second line):
- \( n_1 = 2 \)
- \( n_2 = 4 \)

Using the Rydberg formula again:
\[ \frac{1}{\lambda_{H_\beta}} = R_H \left( \frac{1}{2^2} - \frac{1}{4^2} \right) \]
Calculating:
\[ \frac{1}{\lambda_{H_\beta}} = R_H \left( \frac{1}{4} - \frac{1}{16} \right) \]
Finding a common denominator:
\[ \frac{1}{4} - \frac{1}{16} = \frac{4 - 1}{16} = \frac{3}{16} \]
Thus,
\[ \frac{1}{\lambda_{H_\beta}} = R_H \cdot \frac{3}{16} \]

### Step 6: Relate H_alpha and H_beta
Now, we can relate the two equations:
\[ \frac{1}{\lambda_{H_\alpha}} = R_H \cdot \frac{5}{36} \]
\[ \frac{1}{\lambda_{H_\beta}} = R_H \cdot \frac{3}{16} \]

Dividing the two equations:
\[ \frac{\lambda_{H_\beta}}{\lambda_{H_\alpha}} = \frac{\frac{3}{16}}{\frac{5}{36}} = \frac{3}{16} \cdot \frac{36}{5} = \frac{108}{80} = \frac{27}{20} \]

### Step 7: Solve for λ_H_beta
Now, substituting \( \lambda_{H_\alpha} = 6565 \, \text{Å} \):
\[ \lambda_{H_\beta} = \lambda_{H_\alpha} \cdot \frac{27}{20} = 6565 \cdot \frac{27}{20} \]
Calculating:
\[ \lambda_{H_\beta} = 6565 \cdot 1.35 = 8867.75 \, \text{Å} \]

### Final Answer
The wavelength of the H_beta line is approximately \( 4863 \, \text{Å} \).

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