The wavelength of the first spectral line in the Balmer series of hydrogen atom is 6561 Å. If the wavelength of the spectral line in the Balmer series of singly-ionized helium atom is 1215 when electron jumps from `n_2` to `n_1`, then `n_2` and `n_1`, are

Options

A. `4,2`
B. `5,3`
C. `6,3`
D. `6,2`

Explanation

To solve the problem, we need to determine the values of \( n_2 \) and \( n_1 \) for the spectral line in the Balmer series of singly-ionized helium atom, given that the wavelength is 1215 Å.

### Step-by-step Solution:

1. **Understand the Balmer Series**:
The Balmer series corresponds to transitions of electrons in a hydrogen-like atom where the electron falls to the second energy level (\( n_1 = 2 \)). The first spectral line in the Balmer series for hydrogen corresponds to the transition from \( n_2 = 3 \) to \( n_1 = 2 \).

2. **Identify the Given Values**:
- For the hydrogen atom, the first spectral line has a wavelength of \( \lambda_1 = 6561 \) Å.
- For singly-ionized helium, the wavelength given is \( \lambda_2 = 1215 \) Å.

3. **Use Rydberg's Formula**:
The Rydberg formula for the wavelength of emitted light during electron transitions is given by:
\[ \frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( R \) is the Rydberg constant, \( Z \) is the atomic number, \( n_1 \) is the lower energy level, and \( n_2 \) is the higher energy level.

4. **Apply Rydberg's Formula for Hydrogen**:
For hydrogen (\( Z = 1 \)):
\[ \frac{1}{\lambda_1} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]
Substituting \( \lambda_1 = 6561 \) Å:
\[ \frac{1}{6561} = R \left( \frac{1}{4} - \frac{1}{9} \right) \]
Simplifying:
\[ \frac{1}{4} - \frac{1}{9} = \frac{9 - 4}{36} = \frac{5}{36} \]
Thus,
\[ \frac{1}{6561} = R \cdot \frac{5}{36} \]

5. **Apply Rydberg's Formula for Singly-Ionized Helium**:
For singly-ionized helium (\( Z = 2 \)):
\[ \frac{1}{\lambda_2} = R \cdot 4 \left( \frac{1}{2^2} - \frac{1}{n_2^2} \right) \]
Substituting \( \lambda_2 = 1215 \) Å:
\[ \frac{1}{1215} = R \cdot 4 \left( \frac{1}{4} - \frac{1}{n_2^2} \right) \]

6. **Equate the Two Equations**:
From the hydrogen equation, we can express \( R \):
\[ R = \frac{36}{5 \cdot 6561} \]
Substitute this into the helium equation:
\[ \frac{1}{1215} = \frac{36}{5 \cdot 6561} \cdot 4 \left( \frac{1}{4} - \frac{1}{n_2^2} \right) \]

7. **Solve for \( n_2 \)**:
Rearranging gives:
\[ \frac{1}{1215} = \frac{36 \cdot 4}{5 \cdot 6561 \cdot 4} - \frac{36 \cdot 4}{5 \cdot 6561 \cdot n_2^2} \]
Simplifying:
\[ \frac{1}{1215} = \frac{36}{5 \cdot 6561} - \frac{36}{5 \cdot 6561 \cdot n_2^2} \]
Solving for \( n_2 \) leads us to find that \( n_2 = 4 \).

8. **Conclusion**:
The values of \( n_2 \) and \( n_1 \) are:
\[ n_2 = 4, \quad n_1 = 2 \]

### Final Answer:
\[ (n_2, n_1) = (4, 2) \]