The value of `(n_(2)+n_(1))` and `(n_(2)^(2)-n_(1)^(2))` for `He^(+)` ion in atomic spectrum are 4 and 8 reaspectively . The wave length of emitted photon whwn electron jump from `n_(2)` to `n_(1)` is

Options

A. `(32)/(9)R_(H)`
B. `(9)/(32)R_(H)`
C. `(9)/(32R_(H))`
D. `(32)/(9R_(H))`

Explanation

To solve the problem step-by-step, we will first analyze the given information and then apply the relevant formulas to find the wavelength of the emitted photon when the electron jumps from \( n_2 \) to \( n_1 \).

### Step 1: Understand the Given Information
We are given:
1. \( n_2 + n_1 = 4 \)
2. \( n_2^2 - n_1^2 = 8 \)

### Step 2: Use the Identity for Difference of Squares
We can use the identity \( a^2 - b^2 = (a + b)(a - b) \) to rewrite the second equation:
\[ n_2^2 - n_1^2 = (n_2 + n_1)(n_2 - n_1) \]
Substituting the value from the first equation:
\[ 8 = 4(n_2 - n_1) \]

### Step 3: Solve for \( n_2 - n_1 \)
From the equation \( 8 = 4(n_2 - n_1) \), we can solve for \( n_2 - n_1 \):
\[ n_2 - n_1 = \frac{8}{4} = 2 \]

### Step 4: Set Up the System of Equations
Now we have two equations:
1. \( n_2 + n_1 = 4 \)
2. \( n_2 - n_1 = 2 \)

### Step 5: Solve the System of Equations
We can add these two equations:
\[ (n_2 + n_1) + (n_2 - n_1) = 4 + 2 \]
This simplifies to:
\[ 2n_2 = 6 \implies n_2 = 3 \]

Now substitute \( n_2 \) back into one of the equations to find \( n_1 \):
\[ n_2 + n_1 = 4 \implies 3 + n_1 = 4 \implies n_1 = 1 \]

### Step 6: Use the Rydberg Formula
The wavelength of the emitted photon when the electron jumps from \( n_2 \) to \( n_1 \) can be calculated using the Rydberg formula:
\[ \frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
Where:
- \( R \) is the Rydberg constant (\( R_H \) for hydrogen, but we will use it for helium with \( Z = 2 \)).
- \( Z = 2 \) for helium.

### Step 7: Substitute Values into the Rydberg Formula
Substituting \( n_1 = 1 \), \( n_2 = 3 \), and \( Z = 2 \):
\[ \frac{1}{\lambda} = R \cdot 2^2 \left( \frac{1}{1^2} - \frac{1}{3^2} \right) \]
\[ = R \cdot 4 \left( 1 - \frac{1}{9} \right) \]
\[ = R \cdot 4 \left( \frac{9 - 1}{9} \right) = R \cdot 4 \cdot \frac{8}{9} \]
\[ = \frac{32R}{9} \]

### Step 8: Find the Wavelength
Now, we need to find \( \lambda \):
\[ \lambda = \frac{9}{32R} \]

### Final Answer
Thus, the wavelength of the emitted photon when the electron jumps from \( n_2 \) to \( n_1 \) is:
\[ \lambda = \frac{9}{32R_H} \]