Question

$H^{+},H{e}^{+}$, and $O^{2+}$ all having the same kinetic energy pass through a region in which there is a uniform magnetic field perpendicular to their velocity. The masses of $H^{+},H{e}^{+}$ and $O^{2+}$ are 1, 4, and 16 amu, respectively. Then,

• A. $H^{+}$ will be deflected most
• B. $O^{2+}$ will be deflected most
• C. $H{e}^{+}$ and $O^{2+}$ will be deflected equally
• D. All will be deflected equally

Answer 1

Answer: (A), (C)

$H^{+}$ will be deflected most
$H{e}^{+}$ and $O^{2+}$ will be deflected equally

When the charged particles enter a magnetic field, then a force acts on the particle which will acts as centripetal force.

$qvB=\frac{m{v}^2}{r}\Rightarrow r=\frac{mv}{qB}$ (i)

Now, $E=\frac{1}{2}m{v}^2\Rightarrow v=\frac{\sqrt{2E}}{m}$ (ii)

$\therefore r=\frac{m}{qB}\sqrt{\frac{2E}{m}}=\frac{\sqrt{2Em}}{qB}=\frac{\sqrt{2E}}{B}\times \frac{\sqrt{m}}{q}$

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

$\therefore r_{H^{+}}\propto \frac{\sqrt{1}}{1};p{r}_{H{e}^{+}}\propto \frac{\sqrt{4}}{1};r_{O^{+}}\propto \frac{\sqrt{16\cdot }}{2}$

$\Rightarrow r_{H^{+}}\propto 1;r_{H{e}^{+}}\propto 2;r_{O^{+}}\propto 2$

$\Rightarrow H{e}^{+}$ and $O^{+}$ will be deflected equally.

$H^{+}$ will be deflected the most since its radius is smallest.