According to Boohr's theory the angular momentum of an electron in 5th orbit is :

Options

A. `25(h)/(pi)`
B. `1(h)/(pi)`
C. `10(h)/(pi)`
D. `2.5(h)/(pi)`

Explanation

To calculate the angular momentum of an electron in the 5th orbit according to Bohr's theory, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Formula**: According to Bohr's theory, the angular momentum (L) of an electron in a given orbit is given by the formula:
\[ L = mvr = \frac{n h}{2 \pi} \]
where:
- \(L\) is the angular momentum,
- \(m\) is the mass of the electron,
- \(v\) is the velocity of the electron,
- \(r\) is the radius of the orbit,
- \(n\) is the principal quantum number,
- \(h\) is Planck's constant.

2. **Identify the Principal Quantum Number**: From the question, we know that the electron is in the 5th orbit, which means:
\[ n = 5 \]

3. **Substitute the Values into the Formula**: Now we can substitute the value of \(n\) into the formula for angular momentum:
\[ L = \frac{n h}{2 \pi} = \frac{5 h}{2 \pi} \]

4. **Simplify the Expression**: We can simplify the expression:
\[ L = 2.5 \frac{h}{\pi} \]

5. **Final Result**: Therefore, the angular momentum of the electron in the 5th orbit is:
\[ L = 2.5 \frac{h}{\pi} \]

### Conclusion:
The angular momentum of an electron in the 5th orbit according to Bohr's theory is \(2.5 \frac{h}{\pi}\).

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