The moment of inertia of an electron in `n^(th)` orbit will be

Options

A. `MR^(2)xxn`
B. `(MR^(2)xxn)/(2)`
C. `(2)/(5)MR^(2)xxn`
D. `(2)/(3)MR^(2)xxn`

Explanation

To find the moment of inertia of an electron in the \( n^{th} \) orbit, we can follow these steps:

### Step 1: Understand the Concept of Moment of Inertia
The moment of inertia (\( I \)) of a particle moving in a circular path can be expressed using the formula:
\[ I = m r^2 \]
where \( m \) is the mass of the particle and \( r \) is the radius of the circular path.

### Step 2: Identify the Mass of the Electron
The mass of an electron (\( m \)) is approximately \( 9.11 \times 10^{-31} \) kg.

### Step 3: Determine the Radius of the \( n^{th} \) Orbit
In atomic physics, the radius of the \( n^{th} \) orbit for an electron in a hydrogen atom can be given by:
\[ r_n = n^2 a_0 \]
where \( a_0 \) is the Bohr radius, approximately \( 5.29 \times 10^{-11} \) m.

### Step 4: Substitute the Values into the Moment of Inertia Formula
Now, substituting the radius into the moment of inertia formula:
\[ I_n = m (r_n)^2 = m (n^2 a_0)^2 \]
This simplifies to:
\[ I_n = m n^4 a_0^2 \]

### Step 5: Final Expression for Moment of Inertia
Thus, the moment of inertia of an electron in the \( n^{th} \) orbit can be expressed as:
\[ I_n = m n^4 a_0^2 \]

### Conclusion
The moment of inertia of an electron in the \( n^{th} \) orbit is given by:
\[ I_n = m n^4 a_0^2 \]