Looking for classes? Ksquare Career Institute, Bengaluru →
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 semicircular path of radius $R$ using a magnetic field $B$. If $V$ and $B$ are kept constant, the ratio $\left(\dfrac{\text{charge on the ion}}{\text{mass of the ion}}\right)$ will be proportional to
A
$\dfrac{1}{R^2}$
B
$R^2$
C
$R$
D
$\dfrac{1}{R}$
Detailed Solution
In a mass spectrometer, when the ions are accelerated through a potential $V$: $\dfrac{1}{2}mv^2 = qV$ ...(i)
The magnetic field curves the path of the ions into a semicircular orbit, the magnetic force providing the centripetal force:
$Bqv = \dfrac{mv^2}{R} \Rightarrow v = \dfrac{BqR}{m}$ ...(ii)
Substituting (ii) in (i): $\dfrac{1}{2}m\left(\dfrac{BqR}{m}\right)^2 = qV$
$\dfrac{B^2 q^2 R^2}{2m} = qV$
$\dfrac{q}{m} = \dfrac{2V}{B^2 R^2}$
Since $V$ and $B$ are constants, $\dfrac{q}{m} \propto \dfrac{1}{R^2}$
The magnetic field curves the path of the ions into a semicircular orbit, the magnetic force providing the centripetal force:
$Bqv = \dfrac{mv^2}{R} \Rightarrow v = \dfrac{BqR}{m}$ ...(ii)
Substituting (ii) in (i): $\dfrac{1}{2}m\left(\dfrac{BqR}{m}\right)^2 = qV$
$\dfrac{B^2 q^2 R^2}{2m} = qV$
$\dfrac{q}{m} = \dfrac{2V}{B^2 R^2}$
Since $V$ and $B$ are constants, $\dfrac{q}{m} \propto \dfrac{1}{R^2}$