Question

$H{e}^{+}$ and $O^{2+}$ions (mass of $H{e}^{+}=4amu$ and that of $O^{2+}=16amu$) pass through a region of constant perpendicular magnetic field. If kinetic energy of all the ions is same then

• A. $H{e}^{+}$ions will be deflected more than $O^{2+}$ ions.
• B. $H{e}^{+}$ions will be deflected less than $O^{2+}$ ions
• C. Both the ions will be deflected equally
• D. No ions will be deflected

Answer 1

The correct option is C Both the ions will be deflected equally

$r=\frac{mv}{qB}$, $m=$ mass of charged particle

$r=\frac{\sqrt{2mKE}}{qB}(\because p=\sqrt{2mKE})$

$v=$ velocity of charge $q$
$KE=$ Kinetic energy of particle

Since, $KE$ is same for $H{e}^{+}$ and $O^{2+}$

$\therefore r\propto \frac{\sqrt{m}}{q}$

$\frac{r_{H{e}^{+}}}{r_{O^{2+}}}=\sqrt{\frac{m_{H{e}^{+}}}{m_{O^{2+}}}}\times \frac{q_{O^{2+}}}{q_{H{e}^{+}}}$

$\frac{r_{H{e}^{+}}}{r_{O^{2+}}}=\sqrt{\frac{4}{16}}\times \frac{2}{1}=\frac{1}{1}$

$r_{H{e}^{+}}=r_{O^{2+}}$

So, the deflection is same in both the cases.