Question

Two ions of masses $4\text{amu}$ and $16\text{amu}$ have charges $+2e$ and $+3e$ respectively. These ions pass through the region of constant magnetic field perpendicular to their motion. The kinetic energy of both ions is the same. Then:

• A. lighter ion will be deflected less than heavier ion
• B. no ion will be deflected
• C. lighter ion will be deflected more than heavier ion
• D. both ions will be deflected equally

Answer 1

The correct option is C lighter ion will be deflected more than heavier ion

The radius of circle in which a ions will rotate inside a magnetic field is given by :
$\frac{m{v}^2}{r}=qvB$
$r=\frac{mv}{qB}=\frac{P}{qB}$
We know that :
$P=\sqrt{2mK.E.}$

$\Rightarrow r=\frac{\sqrt{2mK.E.}}{qB}$
From the above expression we can say that :
$r\propto \frac{\sqrt{m}}{q}$
$\Rightarrow \frac{r_1}{r_2}=\frac{\sqrt{4}}{2}\times \frac{3}{\sqrt{16}}$
$\frac{r_1}{r_2}=\frac{3}{4}$
Or, $r_2>r_1$
The bigger the radius of the circular path, the smaller will be the deflection of the ion. Hence, the lighter ion will be deflected more than the heavier ion.