Which among the following has a hydrogen-like spectrum and whose lines have wavelengths four times shorter that those of atomic hydrogen?

Options

A. Helium ion
B. Hydrogen
C. Lithium ion
D. None of these

Answer

To solve the question, we need to determine which atom has a hydrogen-like spectrum and whose spectral lines have wavelengths that are four times shorter than those of atomic hydrogen.

### Step-by-Step Solution:

1. **Understanding the Relationship of Wavelengths**:
- Let \( \lambda_H \) be the wavelength of hydrogen.
- According to the question, the wavelength of the required atom \( \lambda_X \) is four times shorter than that of hydrogen:
\[ \lambda_X = \frac{1}{4} \lambda_H \]

2. **Using Rydberg's Formula**:
- Rydberg's formula for the wavelength of spectral lines is given by:
\[ \frac{1}{\lambda} = R_H Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
- Here, \( R_H \) is the Rydberg constant, \( Z \) is the atomic number, and \( n_1 \) and \( n_2 \) are the principal quantum numbers of the electron transitions.

3. **Applying Rydberg's Formula for Hydrogen**:
- For hydrogen (\( Z = 1 \)):
\[ \frac{1}{\lambda_H} = R_H \cdot 1^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

4. **Applying Rydberg's Formula for the Required Atom**:
- For the hydrogen-like atom (\( Z = Z_X \)):
\[ \frac{1}{\lambda_X} = R_H Z_X^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

5. **Finding the Ratio of Wavelengths**:
- Taking the ratio of the two equations, we have:
\[ \frac{\lambda_X}{\lambda_H} = \frac{R_H \cdot 1^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)}{R_H Z_X^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)} \]
- This simplifies to:
\[ \frac{\lambda_X}{\lambda_H} = \frac{1}{Z_X^2} \]

6. **Substituting the Wavelength Relationship**:
- From step 1, we know:
\[ \frac{1}{4} = \frac{1}{Z_X^2} \]
- Rearranging gives:
\[ Z_X^2 = 4 \]
- Thus, we find:
\[ Z_X = 2 \]

7. **Identifying the Atom**:
- The atomic number \( Z = 2 \) corresponds to helium. However, since we are looking for a hydrogen-like atom, we consider \( \text{He}^+ \) (helium ion), which has one electron and behaves like hydrogen.

8. **Conclusion**:
- The atom with a hydrogen-like spectrum and whose lines have wavelengths four times shorter than those of atomic hydrogen is \( \text{He}^+ \).

### Final Answer:
The required atom is \( \text{He}^+ \) (helium ion).

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