If the wavelength of the first line of the Blamer series of hydrogen atom is `6561Å` , the wavelength of the second line of the series should be
a.13122Å b.3280Å c.4860Å d.2187Å

Options

A. 13122Å
B. 3280Å
C. 4860Å
D. 2187Å

Explanation

To find the wavelength of the second line of the Balmer series for the hydrogen atom, we can follow these steps:

### Step-by-Step Solution:

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

2. **Given Information**:
- Wavelength of the first line (n=3 to n=2): \( \lambda_1 = 6561 \, \text{Å} \)
- For the first line, \( n_1 = 2 \) and \( n_2 = 3 \).

3. **Use the Wavelength Formula**: The formula for the wavelength in the Balmer series is given by:
\[ \frac{1}{\lambda} = R \cdot Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( R \) is the Rydberg constant, and \( Z \) is the atomic number (1 for hydrogen).

4. **Calculate for the First Line**:
- Plugging in the values for the first line:
\[ \frac{1}{6561} = R \cdot 1^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]
\[ \frac{1}{6561} = R \left( \frac{1}{4} - \frac{1}{9} \right) \]
\[ \frac{1}{6561} = R \left( \frac{9 - 4}{36} \right) = R \cdot \frac{5}{36} \]
- Rearranging gives:
\[ R = \frac{36}{5 \cdot 6561} \]

5. **Calculate for the Second Line**:
- For the second line (n=4 to n=2), we have:
\[ \frac{1}{\lambda_2} = R \cdot 1^2 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) \]
\[ \frac{1}{\lambda_2} = R \left( \frac{1}{4} - \frac{1}{16} \right) \]
\[ \frac{1}{\lambda_2} = R \left( \frac{4 - 1}{16} \right) = R \cdot \frac{3}{16} \]

6. **Relate the Two Equations**:
- From the first line, we have:
\[ \frac{1}{6561} = R \cdot \frac{5}{36} \]
- From the second line:
\[ \frac{1}{\lambda_2} = R \cdot \frac{3}{16} \]
- Dividing these two equations gives:
\[ \frac{1/\lambda_2}{1/6561} = \frac{3/16}{5/36} \]
- Simplifying:
\[ \frac{\lambda_2}{6561} = \frac{3 \cdot 36}{5 \cdot 16} \]
\[ \lambda_2 = 6561 \cdot \frac{108}{80} = 6561 \cdot 1.35 = 4860 \, \text{Å} \]

### Final Answer:
The wavelength of the second line of the Balmer series is \( \lambda_2 = 4860 \, \text{Å} \).