The value of orbit angular momentum of an electron in the `3^(rd)` Bohr orbit of hydrogen will be

Options

A. `2.88 xx 10^(-33) kgm^(2) s^(-1)`
B. `7.52 xx 10^(-36) kgm^(2) s^(-1)`
C. `3.16xx 10^(-34) kgm^(2) s^(-1)`
D. `2.91 xx 10^(-35) kgm^(2) s^(-1)`

Explanation

To find the value of the orbital angular momentum of an electron in the third Bohr orbit of hydrogen, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Formula for Orbital Angular Momentum**:
The formula for the orbital angular momentum \( L \) of an electron in the \( n^{th} \) Bohr orbit is given by:
\[ L = \frac{n h}{2\pi} \]
where:
- \( L \) = orbital angular momentum
- \( n \) = principal quantum number (in this case, \( n = 3 \))
- \( h \) = Planck's constant, approximately \( 6.626 \times 10^{-34} \, \text{Js} \)

2. **Substitute the Values**:
For the third Bohr orbit, substitute \( n = 3 \) into the formula:
\[ L = \frac{3 \times h}{2\pi} \]
Now, substituting the value of \( h \):
\[ L = \frac{3 \times 6.626 \times 10^{-34}}{2 \times 3.14} \]

3. **Calculate the Denominator**:
First, calculate \( 2\pi \):
\[ 2\pi \approx 2 \times 3.14 = 6.28 \]

4. **Perform the Multiplication**:
Now, calculate the numerator:
\[ 3 \times 6.626 \times 10^{-34} = 19.878 \times 10^{-34} \]

5. **Divide the Values**:
Now, divide the result from the multiplication by \( 6.28 \):
\[ L = \frac{19.878 \times 10^{-34}}{6.28} \approx 3.17 \times 10^{-34} \, \text{kg m}^2/\text{s} \]

6. **Final Result**:
The value of the orbital angular momentum of an electron in the third Bohr orbit of hydrogen is approximately:
\[ L \approx 3.17 \times 10^{-34} \, \text{kg m}^2/\text{s} \]