What is the wave length of `H_(b)` line Balmer series of hydrogen spectrum ? (R = Rydberg constant)

Options

A. 36/5R
B. 5R/36
C. 3R/16
D. 16/3R

Answer

To find the wavelength of the H-beta line of the Balmer series of the hydrogen spectrum, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Transition Levels**:
The H-beta line corresponds to a transition from n=4 to n=2 in the Balmer series of hydrogen.

2. **Use the Rydberg Formula**:
The Rydberg formula for the wavelength (λ) of light emitted during an electron transition in a hydrogen atom is given by:
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( R \) is the Rydberg constant, \( n_1 \) is the lower energy level, and \( n_2 \) is the higher energy level.

3. **Substitute the Values**:
For the H-beta line:
- \( n_2 = 4 \)
- \( n_1 = 2 \)

Substituting these values into the formula:
\[ \frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{4^2} \right) \]

4. **Calculate the Terms**:
Now calculate \( \frac{1}{2^2} \) and \( \frac{1}{4^2} \):
\[ \frac{1}{2^2} = \frac{1}{4} \quad \text{and} \quad \frac{1}{4^2} = \frac{1}{16} \]
Therefore,
\[ \frac{1}{\lambda} = R \left( \frac{1}{4} - \frac{1}{16} \right) \]

5. **Find a Common Denominator**:
The common denominator for \( \frac{1}{4} \) and \( \frac{1}{16} \) is 16:
\[ \frac{1}{4} = \frac{4}{16} \]
Thus,
\[ \frac{1}{\lambda} = R \left( \frac{4}{16} - \frac{1}{16} \right) = R \left( \frac{3}{16} \right) \]

6. **Solve for λ**:
Rearranging gives us:
\[ \lambda = \frac{16}{3R} \]

### Final Answer:
The wavelength of the H-beta line of the Balmer series of the hydrogen spectrum is:
\[ \lambda = \frac{16}{3R} \]