The wavelength of the first line of Lyman series in hydrogen atom is `1216`. The wavelength of the first line of Lyman series for `10` times ionized sodium atom will be added

Options

A. `0.1 Å`
B. `1000 Å`
C. `100 Å`
D. `10 Å`

Explanation

To solve the problem, we need to find the wavelength of the first line of the Lyman series for a sodium ion that has been ionized 10 times (Na^10+). The given wavelength for the hydrogen atom's first line of the Lyman series is 1216 Å.

### Step-by-Step Solution:

1. **Identify the Wavelength for Hydrogen**:
The wavelength of the first line of the Lyman series for a hydrogen atom (H) is given as:
\[ \lambda_H = 1216 \, \text{Å} \]

2. **Determine the Atomic Number for Sodium**:
The atomic number (Z) of sodium (Na) is 11. When sodium is ionized 10 times, it loses 10 electrons, leaving it with only 1 electron. Therefore, for Na^10+, the effective atomic number remains:
\[ Z = 11 \]

3. **Use the Wavelength Formula**:
The wavelength (λ) for the Lyman series can be related to the atomic number (Z) using the formula:
\[ \lambda \propto \frac{1}{Z^2} \]
This means that the wavelength for the sodium ion can be expressed as:
\[ \lambda_{Na} = \lambda_H \cdot \frac{Z_H^2}{Z_{Na}^2} \]
where \( Z_H = 1 \) for hydrogen and \( Z_{Na} = 11 \).

4. **Substituting Values**:
Now we substitute the values into the equation:
\[ \lambda_{Na} = 1216 \, \text{Å} \cdot \frac{1^2}{11^2} \]
\[ \lambda_{Na} = 1216 \, \text{Å} \cdot \frac{1}{121} \]
\[ \lambda_{Na} = \frac{1216}{121} \, \text{Å} \]
\[ \lambda_{Na} = 10 \, \text{Å} \]

5. **Final Answer**:
The wavelength of the first line of the Lyman series for the 10 times ionized sodium atom is:
\[ \lambda_{Na} = 10 \, \text{Å} \]

### Summary:
The wavelength of the first line of the Lyman series for a 10 times ionized sodium atom is 10 Å.