Find the quantum of the excited state of electrons in `He^(+)` ion which on transition to first excited state emit photons of wavelengths `108.5 nm`. `(R_(H)=1.09678xx10^(7) m^(-1))`

Options

A. `6`
B. `5`
C. `4`
D. `2`

Explanation

To find the quantum of the excited state of electrons in the `He^(+)` ion, we can use the Rydberg formula for hydrogen-like atoms. The steps to solve the problem are as follows:

### Step 1: Understand the Rydberg Formula
The Rydberg formula for the wavelength of emitted or absorbed light during electronic transitions in hydrogen-like atoms is given by:

\[ \frac{1}{\lambda} = R_H \cdot Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

Where:
- \( \lambda \) is the wavelength of the emitted photon,
- \( R_H \) is the Rydberg constant (\( 1.09678 \times 10^7 \, \text{m}^{-1} \)),
- \( Z \) is the atomic number (for \( He^+ \), \( Z = 2 \)),
- \( n_1 \) is the principal quantum number of the lower energy level,
- \( n_2 \) is the principal quantum number of the higher energy level.

### Step 2: Convert Wavelength to Meters
The given wavelength is \( 108.5 \, \text{nm} \). We need to convert this to meters:

\[ \lambda = 108.5 \, \text{nm} = 108.5 \times 10^{-9} \, \text{m} \]

### Step 3: Substitute Values into the Rydberg Formula
For the transition to the first excited state, we have:
- \( n_1 = 1 \) (ground state),
- \( n_2 \) is what we need to find.

Substituting the known values into the Rydberg formula:

\[ \frac{1}{\lambda} = R_H \cdot Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

Substituting \( R_H = 1.09678 \times 10^7 \, \text{m}^{-1} \), \( Z = 2 \), and \( n_1 = 1 \):

\[ \frac{1}{108.5 \times 10^{-9}} = 1.09678 \times 10^7 \cdot 4 \left( 1 - \frac{1}{n_2^2} \right) \]

### Step 4: Calculate \( \frac{1}{\lambda} \)
Calculating \( \frac{1}{\lambda} \):

\[ \frac{1}{\lambda} = \frac{1}{108.5 \times 10^{-9}} \approx 9.21 \times 10^6 \, \text{m}^{-1} \]

### Step 5: Set Up the Equation
Now we can set up the equation:

\[ 9.21 \times 10^6 = 1.09678 \times 10^7 \cdot 4 \left( 1 - \frac{1}{n_2^2} \right) \]

### Step 6: Solve for \( n_2^2 \)
Now simplify and solve for \( n_2^2 \):

\[ 9.21 \times 10^6 = 4.38712 \times 10^7 \left( 1 - \frac{1}{n_2^2} \right) \]

Dividing both sides by \( 4.38712 \times 10^7 \):

\[ \frac{9.21 \times 10^6}{4.38712 \times 10^7} = 1 - \frac{1}{n_2^2} \]

Calculating the left side:

\[ 0.209 = 1 - \frac{1}{n_2^2} \]

Rearranging gives:

\[ \frac{1}{n_2^2} = 1 - 0.209 = 0.791 \]

Thus,

\[ n_2^2 = \frac{1}{0.791} \approx 1.27 \]

### Step 7: Calculate \( n_2 \)
Taking the square root:

\[ n_2 \approx 5 \]

### Final Answer
The quantum number of the excited state of electrons in the `He^(+)` ion is \( n_2 = 5 \).