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

2. **Identify the Principal Quantum Number**:
For the third Bohr orbit, the principal quantum number \( n \) is 3.

3. **Substitute the Values into the Formula**:
Plugging in the values into the formula:
\[ L = 3 \cdot \frac{6.626 \times 10^{-34}}{2\pi} \]

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

5. **Perform the Division**:
Now, calculate \( \frac{h}{2\pi} \):
\[ \frac{6.626 \times 10^{-34}}{6.28} \approx 1.055 \times 10^{-34} \, \text{Js} \]

6. **Multiply by the Principal Quantum Number**:
Now, multiply by \( n \):
\[ L = 3 \cdot 1.055 \times 10^{-34} \approx 3.165 \times 10^{-34} \, \text{Js} \]

7. **Final Answer**:
The value of the orbital angular momentum of an electron in the third Bohr orbit of hydrogen is:
\[ L \approx 3.165 \times 10^{-34} \, \text{kg m}^2/\text{s} \]

### Summary:
The value of the orbital angular momentum of an electron in the third Bohr orbit of hydrogen is approximately \( 3.165 \times 10^{-34} \, \text{kg m}^2/\text{s} \).

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