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 follow semicircular paths of radius $R$ using a magnetic field $B$. If $V$ and $B$ are kept constant, the ratio $(\dfrac{{{\text{charge on the ion}}}}{{{\text{mass of the ion}}}})$ will be proportional to
(A) $\dfrac{1}{R}$
(B) $\dfrac{1}{{{R^2}}}$
(C) ${R^2}$
(D) $R$

Answer 1

Hint Find the energy imparted to the electron when it passes through the electric voltage. This kinetic energy imparted will give $v$ (velocity). After that, we will find the force acting on the particle in the magnetic field. This will have a velocity that will be substituted from the above equation to get the required ratio of $(\dfrac{{{\text{charge on the ion}}}}{{{\text{mass of the ion}}}})$.

Complete step by step answer:
A mass spectrometer is used for measuring the masses of ions.
Initially, the ions are accelerated by an electric potential $V$, which is kept constant.
The electric potential $V$ gives kinetic energy to the ion, which is equivalent to $qV$. Here $q$ is the charge on the ion.
This kinetic energy is equal to the kinetic energy of the particle as, $\dfrac{1}{2}m{v^2} = qV$
In this equation, $m$ is the mass of the ion, and $v$ is the velocity of the ion.
Using this equation of kinetic energy, we get the velocity $v$ as, $v = \sqrt {\dfrac{{2qV}}{m}} $.
This is the velocity that is imparted to an electron when it passes through the given electric potential.
Now, in the magnetic field $B$, the particle is made to follow a semi-circular path of radius $R$.
Here the particle enters with the pre-calculated speed $v = \sqrt {\dfrac{{2qV}}{m}} $
Since the magnetic field is present after the electric potential imparts the required kinetic energy to the particle.
The force on the particle due to the magnetic field $B$ is given by$F = q(\overrightarrow v \times \overrightarrow B )$.
This force is equal to the centripetal force on the particle since any other force is absent in the magnetic field.
Therefore balancing the forces on the particle in the magnetic field we get, $q(\overrightarrow v \times \overrightarrow B ) = \dfrac{{m{v^2}}}{R}$,
Here we will assume that the velocity of the particle when it enters the magnetic field is perpendicular to the magnetic field $B$.
Thus, $qvB\sin \theta = \dfrac{{m{v^2}}}{R}$, where $\theta $ is the angle between the magnetic field and the velocity. Since no value of any angle is given, we assume $\theta $ as ${90^ \circ }$ in this case.
$ \Rightarrow qB\sin {90^ \circ } = \dfrac{{mv}}{R}$,
Substituting the value of $v$, form the equation $v = \sqrt {\dfrac{{2qV}}{m}} $above, we get
$ \Rightarrow qB = \dfrac{m}{R}\sqrt {\dfrac{{2qV}}{m}} $
$ \Rightarrow {q^2}{B^2} = \dfrac{{{m^2}}}{{{R^2}}}(\dfrac{{2qV}}{m})$
$ \Rightarrow q{B^2} = \dfrac{m}{{{R^2}}}(2V)$
$ \Rightarrow \dfrac{q}{m} = \dfrac{{2V}}{{{B^2}{R^2}}}$
$\therefore \dfrac{{{\text{charge on the ion}}}}{{{\text{mass of the ion}}}} = \dfrac{{2V}}{{{B^2}{R^2}}} \propto \dfrac{1}{{{R^2}}}$
As can be seen, the ratio of $(\dfrac{{{\text{charge on the ion}}}}{{{\text{mass of the ion}}}})$ is directly proportional to $\dfrac{1}{{{R^2}}}$.

Thus the option $(B)$ is correct.

Note While considering this problem, we will assume that the particle is encountering only the electric field initially and then the magnetic field. The particle with some charge will not encounter both the fields at the same time. In any case, we will consider the mass of the particle as its rest mass.