The angular momentum of an electron in a Bohr's orbit of `He^(+)` is `3.1652xx10^(-34)` kg-`m^(2)`/sec. What is the wave number in terms of Rydberg constant (R ) of the spectral line emitted when an electron falls from this level to the first excited state.l [ Use h`=6.626xx10^(-34)` Js]

Options

A. (a) `3R`
B. (b) `(5R)/9`
C. (c) `(3R)/(4)`
D. (d) `(8R)/(9)`

Answer

To solve the problem step by step, we will follow the concepts of Bohr's model of the atom and the Rydberg formula for spectral lines.

### Step 1: Understand the given data
We are given:
- Angular momentum \( L = 3.1652 \times 10^{-34} \, \text{kg m}^2/\text{s} \)
- Planck's constant \( h = 6.626 \times 10^{-34} \, \text{Js} \)
- The electron transitions from the third energy level (n=3) to the second energy level (n=2) in the \( He^+ \) ion.

### Step 2: Calculate the principal quantum number \( n \)
According to Bohr's theory, the angular momentum \( L \) of an electron in an orbit is given by:
\[ L = n \frac{h}{2\pi} \]
Rearranging this formula to solve for \( n \):
\[ n = \frac{L \cdot 2\pi}{h} \]
Substituting the known values:
\[ n = \frac{3.1652 \times 10^{-34} \cdot 2\pi}{6.626 \times 10^{-34}} \]
Calculating \( 2\pi \approx 6.2832 \):
\[ n = \frac{3.1652 \times 10^{-34} \cdot 6.2832}{6.626 \times 10^{-34}} \approx 3 \]

### Step 3: Identify the transition levels
Since \( n = 3 \) corresponds to the initial state and the first excited state is \( n = 2 \), the electron transitions from \( n = 3 \) to \( n = 2 \).

### Step 4: Use the Rydberg formula for wave number
The wave number \( \bar{\nu} \) is given by the Rydberg formula:
\[ \bar{\nu} = R \cdot Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
Where:
- \( R \) is the Rydberg constant,
- \( Z \) is the atomic number (for \( He^+ \), \( Z = 2 \)),
- \( n_1 = 2 \) (lower energy level),
- \( n_2 = 3 \) (higher energy level).

### Step 5: Substitute values into the Rydberg formula
Substituting the values into the formula:
\[ \bar{\nu} = R \cdot 2^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]
Calculating:
\[ \bar{\nu} = R \cdot 4 \left( \frac{1}{4} - \frac{1}{9} \right) \]
Finding a common denominator (36):
\[ \bar{\nu} = R \cdot 4 \left( \frac{9 - 4}{36} \right) = R \cdot 4 \cdot \frac{5}{36} \]
Simplifying:
\[ \bar{\nu} = \frac{20}{36} R = \frac{5}{9} R \]

### Final Answer
The wave number in terms of the Rydberg constant \( R \) is:
\[ \bar{\nu} = \frac{5}{9} R \]