The wavelength of `H_(alpha)` line is Balmer series is 6563 Å. Compute the wavelength of `H_(beta)` line of Balmer series.

Answer

To compute the wavelength of the H-beta line of the Balmer series, we will follow these steps:

### Step 1: Understand the Balmer Series
The Balmer series corresponds to the transitions of electrons in a hydrogen atom from higher energy levels (n ≥ 3) to the second energy level (n = 2). The lines in the Balmer series are denoted as H-alpha, H-beta, H-gamma, etc.

### Step 2: Identify the Wavelength of H-alpha
We are given that the wavelength of the H-alpha line (transition from n=3 to n=2) is 6563 Å.

### Step 3: Use the Rydberg Formula
The Rydberg formula for the wavelength of the emitted light during the transition of electrons in hydrogen is given by:
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_2^2} - \frac{1}{n_1^2} \right) \]
where:
- \( R \) is the Rydberg constant (\( R \approx 1.097 \times 10^7 \, \text{m}^{-1} \)),
- \( n_1 \) is the lower energy level,
- \( n_2 \) is the higher energy level.

### Step 4: Calculate H-alpha Wavelength
For the H-alpha line:
- \( n_1 = 2 \)
- \( n_2 = 3 \)

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

### Step 5: Calculate H-beta Wavelength
For the H-beta line:
- \( n_1 = 2 \)
- \( n_2 = 4 \)

Substituting into the Rydberg formula:
\[ \frac{1}{\lambda_{\text{beta}}} = R \left( \frac{1}{2^2} - \frac{1}{4^2} \right) \]
\[ \frac{1}{\lambda_{\text{beta}}} = R \left( \frac{1}{4} - \frac{1}{16} \right) \]
Calculating the right-hand side:
\[ \frac{1}{4} - \frac{1}{16} = \frac{4 - 1}{16} = \frac{3}{16} \]
Thus,
\[ \frac{1}{\lambda_{\text{beta}}} = R \cdot \frac{3}{16} \]

### Step 6: Relate H-alpha and H-beta
From the earlier calculation for H-alpha:
\[ \frac{1}{\lambda_{\text{alpha}}} = R \cdot \frac{5}{36} \]
We can express \( R \) in terms of \( \lambda_{\text{alpha}} \):
\[ R = \frac{36}{5 \lambda_{\text{alpha}}} \]

Substituting this into the equation for \( \lambda_{\text{beta}} \):
\[ \frac{1}{\lambda_{\text{beta}}} = \frac{36}{5 \lambda_{\text{alpha}}} \cdot \frac{3}{16} \]
\[ \frac{1}{\lambda_{\text{beta}}} = \frac{108}{80 \lambda_{\text{alpha}}} \]
\[ \lambda_{\text{beta}} = \frac{80 \lambda_{\text{alpha}}}{108} \]

### Step 7: Substitute the Value of H-alpha
Now substituting \( \lambda_{\text{alpha}} = 6563 \, \text{Å} \):
\[ \lambda_{\text{beta}} = \frac{80 \times 6563}{108} \]
Calculating this gives:
\[ \lambda_{\text{beta}} \approx 6090.74 \, \text{Å} \]

### Final Answer
The wavelength of the H-beta line of the Balmer series is approximately **6091 Å**.

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