The first line of the lyman series in a hydrogen spectrum has a wavelength of 1210 Å. The corresponding line of a hydrogen like atom of `Z=11` is equal to

Options

A. 4000 Ã…
B. 100 Ã…
C. 40 Ã…
D. 10 Ã…

Answer

To solve the problem, we need to find the wavelength of the first line of the Lyman series for a hydrogen-like atom with atomic number \( Z = 11 \). The first line of the Lyman series in hydrogen has a wavelength of \( 1210 \, \text{Å} \).

### Step-by-Step Solution:

1. **Understand the Lyman Series**:
The Lyman series corresponds to transitions from higher energy levels to the first energy level (n=1) in hydrogen. The first line corresponds to the transition from n=2 to n=1.

2. **Use the Rydberg Formula**:
The wavelength of the emitted light can be calculated using the Rydberg formula for hydrogen-like atoms:
\[ \frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( R \) is the Rydberg constant (\( R \approx 1.097 \times 10^7 \, \text{m}^{-1} \)), \( Z \) is the atomic number, \( n_1 \) is the lower energy level, and \( n_2 \) is the higher energy level.

3. **Substituting Values for Hydrogen**:
For hydrogen, the first line of the Lyman series corresponds to \( n_1 = 1 \) and \( n_2 = 2 \):
\[ \frac{1}{\lambda_H} = R \cdot 1^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = R \cdot \frac{3}{4} \]
Given \( \lambda_H = 1210 \, \text{Å} \), we can express this as:
\[ \frac{1}{1210 \times 10^{-10}} = R \cdot \frac{3}{4} \]

4. **Calculate the Wavelength for \( Z = 11 \)**:
Now, substituting \( Z = 11 \) into the Rydberg formula:
\[ \frac{1}{\lambda_{Z=11}} = R \cdot 11^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \cdot 121 \cdot \frac{3}{4} \]
Therefore:
\[ \lambda_{Z=11} = \frac{1}{R \cdot 121 \cdot \frac{3}{4}} \]

5. **Relate the Wavelengths**:
Since we already have \( \frac{1}{\lambda_H} = R \cdot \frac{3}{4} \), we can express \( \lambda_{Z=11} \) in terms of \( \lambda_H \):
\[ \lambda_{Z=11} = \frac{1210 \, \text{Å}}{121} = 10 \, \text{Å} \]

### Final Answer:
The corresponding line of a hydrogen-like atom with \( Z = 11 \) is \( 10 \, \text{Å} \).