A small piece of mass m moves in such a way the P.E. =`-(1)/(2)mkr^(2)`. Where k is a constant and r is the distance of the particle from origin. Assuming Bohr's model of quantization of angular momentum and circular orbit, r is directly proportional to :
(a)`n^(2)`
(b)n
(c)`sqrt(n)`
(d)none of these

Options

A. `n^(2)`
B. n
C. `sqrt(n)`
D. none of these

Answer

To solve the problem, we need to analyze the relationship between the distance \( r \) from the origin and the quantum number \( n \) using the given potential energy expression and Bohr's model.

### Step-by-Step Solution:

1. **Understanding the Potential Energy**:
The potential energy (P.E.) is given as:
\[ P.E. = -\frac{1}{2} m k r^2 \]
where \( m \) is the mass, \( k \) is a constant, and \( r \) is the distance from the origin.

2. **Relating Potential Energy to Kinetic Energy**:
According to classical mechanics, the kinetic energy (K.E.) is equal to half of the potential energy in this context:
\[ K.E. = -\frac{1}{2} P.E. = \frac{1}{4} m k r^2 \]

3. **Using Bohr's Model of Quantization**:
In Bohr's model, the angular momentum \( L \) is quantized and given by:
\[ L = mvr = \frac{n h}{2 \pi} \]
where \( v \) is the velocity of the particle, \( n \) is the principal quantum number, and \( h \) is Planck's constant.

4. **Expressing Velocity in Terms of \( n \)**:
Rearranging the angular momentum equation gives:
\[ v = \frac{n h}{2 \pi m r} \]

5. **Substituting Velocity into Kinetic Energy**:
The kinetic energy can also be expressed in terms of \( v \):
\[ K.E. = \frac{1}{2} m v^2 = \frac{1}{2} m \left(\frac{n h}{2 \pi m r}\right)^2 \]
Simplifying this, we get:
\[ K.E. = \frac{n^2 h^2}{8 \pi^2 m r^2} \]

6. **Equating the Two Expressions for Kinetic Energy**:
From the previous steps, we have two expressions for kinetic energy:
\[ \frac{1}{4} m k r^2 = \frac{n^2 h^2}{8 \pi^2 m r^2} \]

7. **Cross-Multiplying and Rearranging**:
Cross-multiplying gives:
\[ 2 \pi^2 m^2 k r^4 = n^2 h^2 \]
Rearranging for \( r^4 \):
\[ r^4 = \frac{n^2 h^2}{2 \pi^2 m^2 k} \]

8. **Taking the Square Root**:
Taking the square root of both sides gives:
\[ r^2 = \sqrt{\frac{n^2 h^2}{2 \pi^2 m^2 k}} \implies r = \sqrt{\frac{n h}{\sqrt{2} \pi m k}} \]

9. **Identifying the Proportionality**:
From the final expression, we can see that \( r \) is directly proportional to \( \sqrt{n} \):
\[ r \propto \sqrt{n} \]

### Conclusion:
Thus, the correct answer is:
**(c) \( \sqrt{n} \)**