The emission spectra is observed by the consequence of transition of electrons from higher energy state to ground state of `He^(+)` ion. Six different photons are observed during the emission spectra, then what will be the minimum wavelength during the transition?

Options

A. `(4)/(27R_(H))`
B. `(4)/(15R_(H))`
C. `lambda=(15)/(16R_(H))`
D. `(16)/(15R_(H))`

Answer

To solve the problem of finding the minimum wavelength during the transition of electrons in the `He^(+)` ion, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
We know that the emission spectrum is observed due to the transition of electrons from higher energy states to the ground state. We are given that six different photons are observed, which indicates that there are six spectral lines.

2. **Using the Formula for Spectral Lines**:
The number of spectral lines (N) that can be observed during transitions between two energy levels is given by the formula:
\[ N = \frac{(n_2 - n_1)(n_2 - n_1 + 1)}{2} \]
where \( n_1 \) and \( n_2 \) are the principal quantum numbers of the lower and upper energy levels, respectively.

3. **Setting Up the Equation**:
Since we know that \( N = 6 \), we can set up the equation:
\[ 6 = \frac{(n_2 - n_1)(n_2 - n_1 + 1)}{2} \]
Multiplying both sides by 2 gives:
\[ 12 = (n_2 - n_1)(n_2 - n_1 + 1) \]

4. **Finding \( n_1 \) and \( n_2 \)**:
We can test integer values for \( n_1 \) and \( n_2 \) that satisfy this equation. After some trials, we find:
- Let \( n_1 = 1 \) and \( n_2 = 4 \).
- This gives \( (4 - 1)(4 - 1 + 1) = 3 \times 4 = 12 \), which is correct.

5. **Calculating the Minimum Wavelength**:
The formula to calculate the wavelength (\( \lambda \)) for the transition is given by:
\[ \frac{1}{\lambda} = R_H \cdot Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( R_H \) is the Rydberg constant, and \( Z \) is the atomic number of the element. For \( He^{+} \), \( Z = 2 \).

6. **Substituting the Values**:
Substituting \( n_1 = 1 \), \( n_2 = 4 \), and \( Z = 2 \):
\[ \frac{1}{\lambda} = R_H \cdot (2^2) \left( \frac{1}{1^2} - \frac{1}{4^2} \right) \]
\[ = R_H \cdot 4 \left( 1 - \frac{1}{16} \right) \]
\[ = R_H \cdot 4 \left( \frac{15}{16} \right) \]
\[ = \frac{15 R_H}{4} \]

7. **Finding \( \lambda \)**:
Therefore, we have:
\[ \lambda = \frac{4}{15 R_H} \]

### Final Answer:
The minimum wavelength during the transition is:
\[ \lambda = \frac{4}{15 R_H} \]