The angular momentum of an electron in an orbit is quantized because irt is a necessary condition for the compatibility with

Options

A. the wave nature of electron
B. particle nature of electron
C. Paullis exclusion behaviour
D. None of these

Answer

To solve the question regarding the quantization of angular momentum of an electron in an orbit, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Angular Momentum in Bohr's Model**:
- According to Bohr's model of the atom, the angular momentum (L) of an electron in a circular orbit is given by the formula:
\[ L = n \frac{h}{2\pi} \]
where \(n\) is a positive integer (quantum number) and \(h\) is Planck's constant.

2. **Relating Angular Momentum to Linear Momentum**:
- The angular momentum can also be expressed in terms of the linear momentum (p) of the electron:
\[ L = mvr \]
where \(m\) is the mass of the electron, \(v\) is its velocity, and \(r\) is the radius of the orbit.

3. **Equating the Two Expressions for Angular Momentum**:
- By equating the two expressions for angular momentum, we have:
\[ mvr = n \frac{h}{2\pi} \]

4. **Rearranging the Equation**:
- Rearranging gives:
\[ r = \frac{nh}{2\pi mv} \]

5. **Introducing De Broglie Wavelength**:
- The de Broglie wavelength (\(\lambda\)) of the electron is given by:
\[ \lambda = \frac{h}{p} = \frac{h}{mv} \]
- Substituting this into the equation for \(r\), we find:
\[ r = \frac{n\lambda}{2\pi} \]

6. **Conclusion on Quantization**:
- The relationship \(r = \frac{n\lambda}{2\pi}\) indicates that the circumference of the electron's orbit must be an integer multiple of its wavelength, which leads to the quantization of angular momentum. This is a direct consequence of the wave nature of the electron.

7. **Final Answer**:
- Therefore, the quantization of angular momentum of an electron in an orbit is due to its wave nature. The correct option is:
**A) Wave nature of electron**.