Orbits of a particle moving in a circle are such that the perimeter of the orbit equals an integer number of de-Broglie wavelengths of the particle. For a charged particle moving in a plane perpendicular to a magnetic field, the radius of the `n^(th)` orbital will therefore be proportional to:

Options

A. `n^2`
B. `n`
C. `n^(1//2)`
D. `n^(1//4)`

Explanation

To solve the problem, we need to analyze the motion of a charged particle in a magnetic field and relate it to the de Broglie wavelength condition.

### Step-by-Step Solution:

1. **Understanding the de Broglie Wavelength Condition**:
The condition states that the perimeter of the orbit (circumference) is an integer multiple of the de Broglie wavelength. Mathematically, this can be expressed as:
\[ 2\pi r = n \lambda \]
where \( r \) is the radius of the orbit, \( n \) is an integer, and \( \lambda \) is the de Broglie wavelength given by:
\[ \lambda = \frac{h}{p} \]
where \( p \) is the momentum of the particle and \( h \) is Planck's constant.

2. **Relating Momentum to Velocity**:
For a charged particle moving in a magnetic field, the momentum \( p \) can be expressed as:
\[ p = mv \]
where \( m \) is the mass of the particle and \( v \) is its velocity.

3. **Setting Up the Forces**:
The magnetic force acting on the particle is given by:
\[ F_{\text{magnetic}} = qvB \]
where \( q \) is the charge of the particle and \( B \) is the magnetic field strength. This magnetic force provides the necessary centripetal force to keep the particle in circular motion:
\[ F_{\text{centripetal}} = \frac{mv^2}{r} \]

4. **Equating Forces**:
Setting the magnetic force equal to the centripetal force gives:
\[ qvB = \frac{mv^2}{r} \]

5. **Rearranging the Equation**:
Rearranging the above equation to solve for \( r \):
\[ r = \frac{mv}{qB} \]

6. **Substituting Momentum**:
Now substitute \( p = mv \) into the equation for \( r \):
\[ r = \frac{p}{qB} \]

7. **Using the de Broglie Wavelength**:
From the de Broglie wavelength condition, we have:
\[ p = \frac{nh}{2\pi r} \]
Substituting this expression for \( p \) into the equation for \( r \):
\[ r = \frac{nh}{2\pi qBr} \]

8. **Solving for \( r \)**:
Rearranging gives:
\[ r^2 = \frac{nh}{2\pi qB} \]
Therefore, we find that:
\[ r \propto \sqrt{n} \]

### Conclusion:
Thus, the radius of the \( n^{th} \) orbital is proportional to \( \sqrt{n} \):
\[ r \propto \sqrt{n} \]