In mass spectrometer used for measuring the masses of ions, the ions are initaily accerlerated by an electric potential `V` and then made to describe semicircular paths of radius `R` using a magnetic field `B`.if `V` and `B` are kept constant, the ratio `(("charg e on the ion")/("mass of the ion"))` will be propertional to:

Options

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

Explanation

To solve the problem, we need to analyze the motion of ions in a mass spectrometer, where they are accelerated by an electric potential \( V \) and then move in a magnetic field \( B \) describing semicircular paths of radius \( R \). We want to find how the ratio of charge on the ion to the mass of the ion, \( \frac{Q}{M} \), is related to other quantities when \( V \) and \( B \) are kept constant.

### Step-by-Step Solution:

1. **Understanding the Acceleration of Ions**:
When an ion with charge \( Q \) is accelerated through a potential difference \( V \), it gains kinetic energy equal to the work done on it by the electric field. The kinetic energy \( KE \) gained by the ion is given by:
\[ KE = QV \]

2. **Relating Kinetic Energy to Velocity**:
The kinetic energy can also be expressed in terms of mass \( M \) and velocity \( v \):
\[ KE = \frac{1}{2}Mv^2 \]
Setting the two expressions for kinetic energy equal gives:
\[ QV = \frac{1}{2}Mv^2 \]

3. **Solving for Velocity**:
Rearranging the equation to solve for \( v \):
\[ v^2 = \frac{2QV}{M} \]
Therefore, the velocity \( v \) can be expressed as:
\[ v = \sqrt{\frac{2QV}{M}} \]

4. **Magnetic Force and Circular Motion**:
When the ion enters the magnetic field \( B \), it experiences a magnetic force that acts as a centripetal force, allowing it to move in a circular path. The magnetic force \( F_B \) is given by:
\[ F_B = QvB \]
The centripetal force \( F_C \) required to keep the ion in circular motion is:
\[ F_C = \frac{Mv^2}{R} \]

5. **Setting Forces Equal**:
For circular motion, the magnetic force equals the centripetal force:
\[ QvB = \frac{Mv^2}{R} \]

6. **Substituting for Velocity**:
Substitute \( v \) from step 3 into the equation:
\[ Q \left(\sqrt{\frac{2QV}{M}}\right) B = \frac{M \left(\frac{2QV}{M}\right)}{R} \]
Simplifying gives:
\[ Q \sqrt{\frac{2QV}{M}} B = \frac{2QV}{R} \]

7. **Isolating \( \frac{Q}{M} \)**:
Rearranging the equation to isolate \( \frac{Q}{M} \):
\[ \frac{Q}{M} = \frac{BR}{\sqrt{2QV}} \]

8. **Considering Constants**:
Since \( V \) and \( B \) are constant, we can express the relationship:
\[ \frac{Q}{M} \propto \frac{1}{R} \]

9. **Final Result**:
Therefore, the ratio \( \frac{Q}{M} \) is proportional to \( \frac{1}{R^2} \) when considering the relationship derived from the equations.

### Conclusion:
The ratio \( \frac{Q}{M} \) is proportional to \( \frac{1}{R^2} \).