Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

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 $\left(\frac{\text{Charge on the ion}}{\text{Mass of the ion}}\right)$ $\left(\frac{q}{m}\right)$ will be proportional to:

$\frac{1}{R}$

$\frac{1}{R^2}$

$R^2$

$R$

Step-by-Step Solution

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. $qV = \frac{1}{2}mv^2 \implies v = \sqrt{\frac{2qV}{m}}$
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 . $qvB = \frac{mv^2}{R} \implies v = \frac{qBR}{m}$
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 ( $q/m$ ): $\frac{q}{m} = \frac{2V}{B^2 R^2}$
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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