If the series limit of Lyman series for Hydrogen atom is equal to the series limit Balmer series for a hydorgen like atom, then atomic number of this hydrogen-like atom will be

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem, we need to find the atomic number \( Z \) of a hydrogen-like atom such that the series limit of the Lyman series for hydrogen is equal to the series limit of the Balmer series for that hydrogen-like atom.

### Step-by-Step Solution:

1. **Understanding Series Limit**:
The series limit refers to the shortest wavelength in a spectral series. For the Lyman series, the transition occurs from \( n_2 = \infty \) to \( n_1 = 1 \).

2. **Lyman Series for Hydrogen**:
The formula for the wavelength \( \lambda \) in terms of the Rydberg constant \( R \) for hydrogen is given by:
\[ \frac{1}{\lambda} = R \cdot Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
For the Lyman series:
- \( n_1 = 1 \)
- \( n_2 = \infty \)

Substituting these values:
\[ \frac{1}{\lambda_H} = R \cdot 1^2 \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = R \]
Thus, we have:
\[ \lambda_H = \frac{1}{R} \]

3. **Balmer Series for Hydrogen-like Atom**:
For the Balmer series, the transitions occur from \( n_2 = \infty \) to \( n_1 = 2 \). Using the same formula:
\[ \frac{1}{\lambda_Z} = R \cdot Z^2 \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) = R \cdot Z^2 \cdot \frac{1}{4} \]
Thus, we have:
\[ \lambda_Z = \frac{4}{R Z^2} \]

4. **Setting the Series Limits Equal**:
According to the problem, the series limit of the Lyman series for hydrogen equals the series limit of the Balmer series for the hydrogen-like atom:
\[ \lambda_H = \lambda_Z \]
Substituting the expressions we derived:
\[ \frac{1}{R} = \frac{4}{R Z^2} \]

5. **Solving for \( Z \)**:
Cancel \( R \) from both sides (assuming \( R \neq 0 \)):
\[ 1 = \frac{4}{Z^2} \]
Rearranging gives:
\[ Z^2 = 4 \]
Taking the square root:
\[ Z = 2 \]

### Final Answer:
The atomic number \( Z \) of the hydrogen-like atom is \( 2 \).

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