Electronic transition in `He^(+)` ion takes from `n_(2) " to " n_(1)` shell such that :
`2n_(2)+3n_(1)=18`
`2n_(2)-3n_(1)=6`
What will be the total number of photons emitted when electrons transit to `n_(1)` shell?

Options

A. 21
B. 15
C. 20
D. 10

Answer

To solve the problem regarding the electronic transition in the `He^(+)` ion, we need to follow these steps:

### Step-by-Step Solution

1. **Set Up the Equations**: We have two equations based on the problem statement:
\[ 2n_2 + 3n_1 = 18 \quad \text{(1)} \]
\[ 2n_2 - 3n_1 = 6 \quad \text{(2)} \]

2. **Add the Equations**: To eliminate one of the variables, we can add equations (1) and (2):
\[ (2n_2 + 3n_1) + (2n_2 - 3n_1) = 18 + 6 \]
This simplifies to:
\[ 4n_2 = 24 \]
Therefore,
\[ n_2 = \frac{24}{4} = 6 \]

3. **Substitute to Find \(n_1\)**: Now that we have \(n_2\), we can substitute it back into one of the original equations to find \(n_1\). We'll use equation (1):
\[ 2(6) + 3n_1 = 18 \]
Simplifying this gives:
\[ 12 + 3n_1 = 18 \]
Rearranging gives:
\[ 3n_1 = 18 - 12 = 6 \]
Thus,
\[ n_1 = \frac{6}{3} = 2 \]

4. **Determine the Number of Photons Emitted**: The number of photons emitted during the transition from \(n_2\) to \(n_1\) can be calculated using the formula:
\[ \text{Number of photons} = \frac{\Delta n (\Delta n + 1)}{2} \]
where \(\Delta n = n_2 - n_1\). Here, we have:
\[ \Delta n = 6 - 2 = 4 \]
Substituting this value into the formula gives:
\[ \text{Number of photons} = \frac{4(4 + 1)}{2} = \frac{4 \times 5}{2} = \frac{20}{2} = 10 \]

5. **Final Answer**: Therefore, the total number of photons emitted when electrons transit to the \(n_1\) shell is:
\[ \boxed{10} \]