In hydrogen spectrum `L_(alpha)` line arises due to transition of electron from the orbit n=3 to the orbit n=2. In the spectrum of singly ionized helium there is a line having the same wavelength as that of the `L_(alpha)` line. This is due to the transition of electron from the orbit:

Options

A. n = 3 to n = 2
B. n = 4 to n = 2
C. n = 5 to n = 3
D. n = 6 to n = 4

Answer

To solve the problem, we need to determine the transition of the electron in singly ionized helium that corresponds to the same wavelength as the L-alpha line in hydrogen. The L-alpha line in hydrogen arises from the transition of an electron from the orbit n=3 to n=2.

### Step-by-Step Solution:

1. **Understand the Transition in Hydrogen:**
The L-alpha line in hydrogen corresponds to the transition of an electron from n=3 to n=2. We can use the Rydberg formula to find the wavelength of this transition:
\[ \frac{1}{\lambda} = R \cdot Z^2 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \]
where \( R \) is the Rydberg constant, \( Z \) is the atomic number (1 for hydrogen), \( n_f \) is the final orbit (2), and \( n_i \) is the initial orbit (3).

2. **Calculate the Wavelength for Hydrogen:**
Substituting the values into the formula:
\[ \frac{1}{\lambda} = R \cdot 1^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) \]
Finding a common denominator (36):
\[ \frac{1}{\lambda} = R \left( \frac{9}{36} - \frac{4}{36} \right) = R \cdot \frac{5}{36} \]
Therefore, we can express the wavelength as:
\[ \lambda = \frac{36}{5R} \]

3. **Transition in Singly Ionized Helium:**
For singly ionized helium (He\(^+\)), the atomic number \( Z = 2 \). We need to find the transition that gives the same wavelength. The formula becomes:
\[ \frac{1}{\lambda} = R \cdot 2^2 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) = 4R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \]
Setting this equal to the previous wavelength:
\[ \frac{5}{36R} = 4R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \]

4. **Equating and Solving for \( n_f \) and \( n_i \):**
Rearranging gives:
\[ \frac{1}{n_f^2} - \frac{1}{n_i^2} = \frac{5}{144} \]
We can try different pairs of \( n_f \) and \( n_i \) to satisfy this equation.

5. **Testing Values:**
Let's test \( n_f = 4 \) and \( n_i = 6 \):
\[ \frac{1}{4^2} - \frac{1}{6^2} = \frac{1}{16} - \frac{1}{36} \]
Finding a common denominator (144):
\[ \frac{9}{144} - \frac{4}{144} = \frac{5}{144} \]
This satisfies our equation.

### Conclusion:
The transition in singly ionized helium that corresponds to the same wavelength as the L-alpha line in hydrogen is from the orbit \( n=6 \) to \( n=4 \).