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 semicircular paths of radius $R$ using a magnetic field $B$. If $V$ and $B$ are kept constant, the ratio $\left(\frac{\text { charge on the ion }}{\text { mass of the ion }}\right)$ will be proportional to

• A. $ \frac{1}{R} $
• B. $ \frac{1}{R^{2}} $
• C. $R^{2}$
• D. $ R $

Answer 1

The Correct Option is B

The radius of the orbit in which ions moving is determined by the relation as given below
$\frac{m v^{2}}{R}=q v B$
where $m$ is the mass, $v$ is velocity, $q$ is charge of ion and $B$ is the flux density of the magnetic field, so that $q v B$ is the magnetic force acting on the ion, and $\frac{m v^{2}}{R}$ is the centripetal force on the ion moving in a curved path of radius $R$.
The angular frequency of rotation of the ions about the vertical field $B$ is given by
$\omega=\frac{v}{R}=\frac{q B}{m}=2 \pi v$
where $v$ is frequency.
Energy of ion is given by
$E =\frac{1}{2} m v^{2}$
$=\frac{1}{2} m(R \omega)^{2}$
$=\frac{1}{2} m R^{2} B^{2} \frac{q^{2}}{m^{2}}$
or $E =\frac{1}{2} \frac{R^{2} B^{2} q^{2}}{m}$ ...(i)
If ions are accelerated by electric potential $V$, then energy attained by ions
$E=q V$ ...(ii)
From Eqs. (i) and (ii), we get
$q V =\frac{1}{2} \frac{R^{2} B^{2} q^{2}}{m}$
or $\frac{q}{m} =\frac{2 V}{R^{2} B^{2}}$
If $V$ and $B$ are kept constant, then
$\frac{q}{m} \propto \frac{1}{R^{2}}$