In the spectrum of singly ionized helium , the wavelength of a line abserved is almost the same as the first line of Balmer series of hydrogen . It is due to transition of electron from `n_(1) = 6 to n_(2) = ^(''**'')`. What is the value of `(''**'')`.

Explanation

To solve the problem, we need to find the value of \( n_2 \) for the transition of an electron in singly ionized helium that produces a wavelength similar to the first line of the Balmer series of hydrogen.

### Step-by-Step Solution:

1. **Understand the Balmer Series for Hydrogen:**
The first line of the Balmer series corresponds to the transition from \( n_2 = 3 \) to \( n_1 = 2 \). The formula for the wavelength in the Balmer series is given by:
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( R \) is the Rydberg constant.

2. **Calculate the Wavelength for the First Line of Balmer Series:**
For the first line (transition from \( n_2 = 3 \) to \( n_1 = 2 \)):
\[ \frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]
Simplifying this:
\[ \frac{1}{\lambda} = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{9 - 4}{36} \right) = R \left( \frac{5}{36} \right) \]
Thus, we have:
\[ \lambda = \frac{36}{5R} \]

3. **Apply the Formula for Singly Ionized Helium:**
For singly ionized helium, the atomic number \( Z = 2 \). The formula for the wavelength in this case is:
\[ \frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
Here, \( n_1 = 6 \) (as given) and \( Z = 2 \):
\[ \frac{1}{\lambda} = R \cdot 2^2 \left( \frac{1}{6^2} - \frac{1}{n_2^2} \right) = 4R \left( \frac{1}{36} - \frac{1}{n_2^2} \right) \]

4. **Set the Two Wavelength Equations Equal:**
Since the wavelengths are almost the same:
\[ \frac{5}{36}R = 4R \left( \frac{1}{36} - \frac{1}{n_2^2} \right) \]
Cancel \( R \) from both sides:
\[ \frac{5}{36} = 4 \left( \frac{1}{36} - \frac{1}{n_2^2} \right) \]

5. **Simplify and Solve for \( n_2^2 \):**
Multiply both sides by 36:
\[ 5 = 4 \left( 1 - \frac{36}{n_2^2} \right) \]
Expanding gives:
\[ 5 = 4 - \frac{144}{n_2^2} \]
Rearranging gives:
\[ 1 = -\frac{144}{n_2^2} \implies \frac{144}{n_2^2} = 1 \implies n_2^2 = 144 \]
Therefore:
\[ n_2 = 12 \]

### Final Answer:
The value of \( n_2 \) is **12**.