In a hypothetical system , a partical of mass `m` and charge `-3 q` is moving around a very heavy partical chaRGE `q`. Assume that Bohr's model is applicable to this system , then velocuity of mass `m` in the first orbit is

Options

A. `(3q^(2))/(2epsi_(0)h)`
B. `(3q^(2))/(4epsi_(0)h)`
C. `(3q)/(2epsi_(0)h)`
D. `(3q)/(4epsi_(0)h)`

Answer

To find the velocity of a particle of mass \( m \) and charge \( -3q \) moving around a very heavy particle with charge \( q \) in the first orbit, we can use the principles derived from Bohr's model of the atom.

### Step-by-Step Solution:

1. **Understanding the System**:
We have a particle with charge \( -3q \) moving in a circular orbit around a heavy particle with charge \( q \). According to Coulomb's law, the electrostatic force provides the necessary centripetal force for circular motion.

2. **Centripetal Force and Electrostatic Force**:
The electrostatic force \( F \) between the two charges is given by:
\[ F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} = \frac{k \cdot |(-3q) \cdot q|}{r^2} = \frac{3kq^2}{r^2} \]
where \( k \) is Coulomb's constant and \( r \) is the radius of the orbit.

3. **Centripetal Force Requirement**:
The centripetal force required to keep the particle in circular motion is given by:
\[ F_c = \frac{mv^2}{r} \]
where \( m \) is the mass of the particle and \( v \) is its velocity.

4. **Setting Forces Equal**:
For the particle to maintain its circular motion, the centripetal force must equal the electrostatic force:
\[ \frac{mv^2}{r} = \frac{3kq^2}{r^2} \]

5. **Rearranging the Equation**:
We can rearrange this equation to find \( v^2 \):
\[ mv^2 = \frac{3kq^2}{r} \]
\[ v^2 = \frac{3kq^2}{mr} \]

6. **Using Bohr's Quantization Condition**:
According to Bohr's model, the angular momentum \( L \) is quantized:
\[ L = n \frac{h}{2\pi} \]
For the first orbit, \( n = 1 \):
\[ L = \frac{h}{2\pi} \]
The angular momentum can also be expressed as:
\[ L = mvr \]
Setting these equal gives:
\[ mvr = \frac{h}{2\pi} \]
From this, we can express \( v \):
\[ v = \frac{h}{2\pi mr} \]

7. **Substituting \( v \) into the Force Equation**:
Now, substitute \( v \) back into the equation for \( v^2 \):
\[ \left(\frac{h}{2\pi mr}\right)^2 = \frac{3kq^2}{mr} \]
Simplifying gives:
\[ \frac{h^2}{4\pi^2 m^2 r^2} = \frac{3kq^2}{mr} \]
Multiplying both sides by \( 4\pi^2 m^2 r^2 \):
\[ h^2 = 12\pi^2 kq^2 m r \]

8. **Solving for \( v \)**:
Now we can solve for \( v \) using the relationship we derived earlier:
\[ v = \sqrt{\frac{3kq^2}{mr}} \]
Substituting \( r \) from the angular momentum equation gives us the final expression for the velocity.

### Final Expression:
After simplifying, we find:
\[ v = \frac{3q^2}{2\pi \epsilon_0 h} \]

### Conclusion:
The velocity of the particle in the first orbit is given by:
\[ v = \frac{3q^2}{2\pi \epsilon_0 h} \]