Question

A proton and a deuteron both having the same kinetic energy, enter perpendicularly into a uniform magnetic field $B$. For motion of proton and deuteron on circular path of radius $r_{P}$ and $r_{d}$ respectively, the correct statement is

• A. $r_{d}=r_{P} \sqrt{2}$
• B. $r_{d} \frac{r_{P}}{\sqrt{2}}$
• C. $r_{d}=r_{P}$
• D. $r_{d}=2 r_{P}$

Answer 1

Answer: (A)

Radius of circular path $r=\frac{m v}{q B}$...(i)
and kinetic energy $K E=\frac{1}{2} m v^{2}$
or $v=\sqrt{\frac{2 K E}{m}}$
putting the value of $v$ in Eq. (i)
$r=\frac{m}{q B} \sqrt{\frac{2 K E}{m}}=\frac{\sqrt{2 m K E}}{q B} \ldots$ (ii)
For proton $\left({ }_{1} H^{1}\right)$ and deuteron $\left({ }_{1} H^{2}\right)$
electronic charge are same and both have same kinetic energy.
$\therefore $ By Eq. (ii) $r \propto \sqrt{m}$
or $\frac{r_{d}}{r_{p}}=\sqrt{\frac{m_{d}}{m_{p}}}$
$\because m_{d}=2 m_{p}$
$\therefore \frac{r_{d}}{r_{p}}=\sqrt{\frac{2 m_{p}}{m_{p}}}$
$\Rightarrow r_{d}=\sqrt{2} r_{p}$