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 Å`
B. `100 Å`
C. `40 Å`
D. `10 Å`

Explanation

To solve the problem of finding the corresponding line of a hydrogen-like atom with atomic number \( Z = 11 \) given that the first line of the Lyman series in hydrogen has a wavelength of \( 1210 \, \text{Å} \), we will follow these steps:

### Step 1: Understand the Lyman Series
The Lyman series corresponds to transitions of electrons in a hydrogen atom from higher energy levels to the first energy level (n=1). The first line of the Lyman series corresponds to the transition from \( n = 2 \) to \( n = 1 \).

### Step 2: Use the Rydberg Formula
The wavelength \( \lambda \) of the spectral lines in hydrogen-like atoms can be calculated using the Rydberg formula:
\[ \frac{1}{\lambda} = R \cdot Z^2 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \]
where:
- \( R \) is the Rydberg constant (\( R \approx 1.097 \times 10^7 \, \text{m}^{-1} \)),
- \( Z \) is the atomic number,
- \( n_f \) is the final energy level,
- \( n_i \) is the initial energy level.

### Step 3: Identify the Values
For hydrogen (\( Z = 1 \)):
- The first line of the Lyman series corresponds to \( n_f = 1 \) and \( n_i = 2 \).
- The wavelength for hydrogen is given as \( \lambda = 1210 \, \text{Å} \).

For the hydrogen-like atom with \( Z = 11 \):
- We will use the same \( n_f = 1 \) and \( n_i = 2 \).

### Step 4: Relate the Wavelengths
Since the wavelength is inversely proportional to \( Z^2 \), we can set up the following relationship:
\[ \frac{\lambda'}{\lambda} = \frac{Z^2}{Z_0^2} \]
where \( \lambda' \) is the wavelength for the hydrogen-like atom, \( Z_0 = 1 \) for hydrogen, and \( Z = 11 \) for the hydrogen-like atom.

### Step 5: Substitute the Values
Substituting the known values:
\[ \frac{\lambda'}{1210 \, \text{Å}} = \frac{11^2}{1^2} \]
\[ \frac{\lambda'}{1210 \, \text{Å}} = 121 \]
Thus,
\[ \lambda' = 1210 \, \text{Å} \times \frac{1}{121} \]
Calculating \( \lambda' \):
\[ \lambda' = \frac{1210}{121} = 10 \, \text{Å} \]

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

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