In a mass spectrometer used for measuring the masses of ions, the ions are initially accelerated by an electri

Question

In a mass spectrometer used for measuring the masses of ions, the ions are initially accelerated by an electric potential $V$ and then made to describe semi-circular paths of radius $R$ using a magnetic field $B$. If $V$ and $B$ are kept constant, the ratio of $\left(\frac{	ext{Charge on the ion}}{	ext{Mass of the ion}}ight)$ $\left(\frac{q}{m}ight)$ will be proportional to:

Accepted Answer

1. **Acceleration by Electric Field:** When an ion with charge $q$ and mass $m$ is accelerated from rest through a potential difference $V$, the work done by the electric field equals the gain in kinetic energy. 
   $$qV = \frac{1}{2}mv^2 \implies v = \sqrt{\frac{2qV}{m}}$$ 
2. **Motion in Magnetic Field:** Upon entering a uniform magnetic field $B$ perpendicular to its velocity, the Lorentz force provides the necessary centripetal force for circular motion .
   $$qvB = \frac{mv^2}{R} \implies v = \frac{qBR}{m}$$ 
3. **Relating Specific Charge (q/m) to R:** Substitute the expression for velocity ($v$) from the magnetic field equation into the energy equation:
   $$qV = \frac{1}{2}m \left( \frac{qBR}{m} \right)^2$$
   $$qV = \frac{1}{2}m \frac{q^2 B^2 R^2}{m^2}$$
   $$V = \frac{1}{2} \frac{q B^2 R^2}{m}$$
   Rearranging to find the charge-to-mass ratio ($q/m$):
   $$\frac{q}{m} = \frac{2V}{B^2 R^2}$$ 
4. **Conclusion:** Since $V$ and $B$ are constant, the specific charge is inversely proportional to the square of the radius .
   $$\frac{q}{m} \propto \frac{1}{R^2}$$

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