Question

A proton and a deuteron are both accelerated through the same potential difference and enter in a magnetic field perpendicular to the direction of the field. If the deuteron follows a path of radius $R$, assuming the neutron and proton masses are nearly equal, the radius of the proton’s path will be

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

Answer 1

Answer: (B)

As charge on both proton and deuteron is same i.e. $'e'$
Energy acquired by both, $E = eV$ For Deuteron.
Kinetic energy, $\frac{1}{2}mV^{2}-eV$
[$V$ is the potential difference]
$v=\sqrt{\frac{2eV}{m_{d}}}$
But $m_{d} = 2m$
Therefore, $v = \sqrt{\frac{2eV}{2m}}=\sqrt{\frac{eV}{m}}$
Radius of path, $R=\frac{mv}{eB}$
Substituting value of $'v'$ we get
$R=\frac{2m\sqrt{\frac{ev}{m}}}{eB}$
$\frac{R}{2}=\frac{m\sqrt{\frac{ev}{m}}}{eB} \,...\left(i\right)$
For proton :
$\frac{1}{2}mV^{2}=eV$
$V=\sqrt{\frac{2eV}{m}}$
Radius of path, $R'=\frac{mV}{eB}=\frac{m\sqrt{\frac{2eV}{m}}}{eB}$
$R'=\sqrt{2}\times\frac{R}{2}$ [From eq. (i)]
$R'=\frac{R}{\sqrt{2}}$