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 $\left(\frac{chargeontheion}{massoftheion}\right)$ 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 $\frac{1}{R^2}$

The radius of the orbit in which ions moving is determined by the relation as given below
$\frac{m{v}^2}{R}=qvB$
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 $qvB$ 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
$\omega =\frac{v}{R}=\frac{qB}{m}=2\pi v$
where $v$ is frequency.
Energy of ion is given by
$E=\frac{1}{2}m{v}^2=\frac{1}{2}m{\left(R\omega \right)}^2$
$=\frac{1}{2}m{R}^2{B}^2\frac{q^2}{m^2}$
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
$E=qV$ .....(ii)
From equations (i) and (ii), we get
$qV=\frac{1}{2}\frac{R^2{B}^2{q}^2}{m}$
or $\frac{q}{m}=\frac{2V}{R^2{B}^2}$
If $V$ and $B$ are kept constant, then
$\frac{q}{m}\propto \frac{1}{R^2}$