The angular momentum of an electron present in the excited state of hydrogen is `1.5h//pi`. The electron is present in

Options

A. Third orbit
B. Second orbit
C. Fourth orbit
D. Fifth orbit

Answer

To solve the problem regarding the angular momentum of an electron in an excited state of hydrogen, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Formula for Angular Momentum**:
According to Bohr's model of the hydrogen atom, the angular momentum (L) of an electron in an orbit is given by the formula:
\[ L = n \frac{h}{2\pi} \]
where \( n \) is the principal quantum number (which indicates the orbit), and \( h \) is Planck's constant.

2. **Set Up the Equation**:
The problem states that the angular momentum of the electron is \( 1.5 \frac{h}{\pi} \). We can equate this to the formula for angular momentum:
\[ n \frac{h}{2\pi} = 1.5 \frac{h}{\pi} \]

3. **Simplify the Equation**:
To simplify, we can multiply both sides of the equation by \( 2\pi \) to eliminate the fractions:
\[ n h = 3h \]

4. **Solve for \( n \)**:
Now, we can divide both sides by \( h \) (assuming \( h \neq 0 \)):
\[ n = 3 \]

5. **Determine the Orbit**:
The value of \( n \) indicates the principal quantum number, which corresponds to the orbit of the electron. Since \( n = 3 \), this means the electron is present in the **third orbit**.

6. **Select the Correct Option**:
From the options given:
- First option: third orbit
- Second option: second orbit
- Third option: fourth orbit
- Fourth option: fifth orbit

The correct answer is the **first option**, which is the third orbit.

### Final Answer:
The electron is present in the **third orbit**.

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