If `l_(3) and l_(2)` represent angular momenta of an orbiting electron in III and II Bohr orbits

Options

A. A) `3:2`
B. B) `9:4`
C. C) `2:3`
D. D) `4:9`

Answer

To solve the problem regarding the angular momenta of an orbiting electron in III and II Bohr orbits, we will follow these steps:

### Step-by-Step Solution:

1. **Understanding Angular Momentum in Bohr's Model**:
According to Bohr's model, the angular momentum \( l_n \) of an electron in the nth orbit is given by the formula:
\[ l_n = n \hbar \]
where \( n \) is the principal quantum number (the orbit number) and \( \hbar \) is the reduced Planck's constant.

2. **Calculate Angular Momentum for III Bohr Orbit**:
For the III Bohr orbit (where \( n = 3 \)):
\[ l_3 = 3 \hbar \]

3. **Calculate Angular Momentum for II Bohr Orbit**:
For the II Bohr orbit (where \( n = 2 \)):
\[ l_2 = 2 \hbar \]

4. **Finding the Ratio of Angular Momenta**:
Now, we need to find the ratio of \( l_3 \) to \( l_2 \):
\[ \frac{l_3}{l_2} = \frac{3 \hbar}{2 \hbar} \]

5. **Simplifying the Ratio**:
The \( \hbar \) cancels out:
\[ \frac{l_3}{l_2} = \frac{3}{2} \]

6. **Final Result**:
Therefore, the ratio of the angular momenta \( l_3 \) to \( l_2 \) is:
\[ l_3 : l_2 = 3 : 2 \]