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 $\left(\frac{Chargeontheion}{massoftheion}\right)$ will be proportional to

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

Answer 1

Answer: (B)

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

The centripetal force is provided by the magnetic force.
i.e., $\frac{m{v}^2}{R}=qvB.......\left(1\right)$
Where m=mass of the ion, v=velocity, q=charge of ion, B= flux density of the magnetic field.
We have, $v=R\omega$
or $\omega =\frac{v}{R}=\frac{qB}{m}$ (From $\left(1\right)$)
Energy of ion is given by,
$E=\frac{1}{2}m{v}^2=\frac{1}{2}m(R\omega {)}^2=\frac{1}{2}m{R}^2\frac{q^2{B}^2}{m^2}$
or $E=\frac{1}{2}\frac{R^2{B}^2{q}^2}{m}.......\left(2\right)$
If ions are accelerated by electric potential $V$, the energy attained by ions,
$E=qV$ ..........$\left(3\right)$
From eqns $\left(2\right)$ and $\left(3\right)$
$qV=\frac{1}{2}\frac{R^2{B}^2{q}^2}{m}$ or $\left(\frac{q}{m}\right)=\frac{2V}{R^2{B}^2}$
i.e., $\left(\frac{q}{m}\right)\propto \frac{1}{R^2}$ (If $V$ and $B$ are const.)