If one were to apply Bohr model to a particle of mass `'m'` and charge `'q'` moving in a plane under the influence of a mgentic filed `'B'`, the energy of the cahrged particle in the `n^(th)` level will be :-

Options

A. `n((hqB)/(2pim))`
B. `n((hqB)/(4pim))`
C. `n((hqB)/(8pim))`
D. `n((hqB)/(pim))`

Explanation

To solve the problem of finding the energy of a charged particle in the nth level under the influence of a magnetic field using the Bohr model, we can follow these steps:

### Step 1: Understanding the Forces
A charged particle with mass \( m \) and charge \( q \) moving in a magnetic field \( B \) experiences a magnetic force given by:
\[ F_{\text{magnetic}} = qvB \]
This magnetic force acts as the centripetal force required to keep the particle in circular motion:
\[ F_{\text{centripetal}} = \frac{mv^2}{r} \]

### Step 2: Setting the Forces Equal
Since the magnetic force provides the centripetal force, we can set the two forces equal:
\[ qvB = \frac{mv^2}{r} \]

### Step 3: Solving for Velocity
Rearranging the equation gives us:
\[ qvB r = mv^2 \]
Dividing both sides by \( v \) (assuming \( v \neq 0 \)):
\[ qBr = mv \]
Thus, we can express the velocity \( v \) as:
\[ v = \frac{qBr}{m} \]

### Step 4: Applying Bohr's Quantization Condition
According to the Bohr model, the angular momentum \( L \) is quantized:
\[ L = mvr = n\frac{h}{2\pi} \]
Substituting for \( v \):
\[ m \left(\frac{qBr}{m}\right) r = n\frac{h}{2\pi} \]
This simplifies to:
\[ qBr^2 = n\frac{h}{2\pi} \]

### Step 5: Finding the Potential Energy
The potential energy \( U \) in a magnetic field can be expressed as:
\[ U = -\mu \cdot B \]
where \( \mu \) is the magnetic moment. The magnetic moment \( \mu \) for a current loop is given by:
\[ \mu = I \cdot A \]
For a circular path, the area \( A = \pi r^2 \) and the current \( I = \frac{q}{T} \) where \( T \) is the period of motion. The frequency \( f = \frac{1}{T} \).

Thus, we can express the potential energy as:
\[ U = -\frac{q}{T} \cdot \pi r^2 B \]
Substituting \( T = \frac{2\pi r}{v} \) gives:
\[ U = -qf \pi r^2 B \]

### Step 6: Total Energy
The total energy \( E \) of the particle is the sum of its kinetic energy \( K \) and potential energy \( U \):
\[ E = K + U \]
The kinetic energy \( K \) is given by:
\[ K = \frac{1}{2} mv^2 = \frac{1}{2} m \left(\frac{qBr}{m}\right)^2 = \frac{q^2 B^2 r^2}{2m} \]
Thus, the total energy becomes:
\[ E = \frac{q^2 B^2 r^2}{2m} - qf \pi r^2 B \]

### Step 7: Substituting for \( r \)
From the earlier equation \( qBr^2 = n\frac{h}{2\pi} \), we can express \( r^2 \) in terms of \( n \):
\[ r^2 = \frac{n h}{2\pi qB} \]
Substituting this back into the energy equation gives us the final expression for the energy of the charged particle in the nth level:
\[ E = n \frac{qBh}{2\pi m} \]

### Final Answer
Thus, the energy of the charged particle in the nth level is:
\[ E_n = n \frac{qBh}{2\pi m} \]