b
As charge on both proton and deuteron is same i.e. \('e'\)

Energy acquired by both, \(E=e V\) For Deuteron.

Kinetic energy, \(\frac{1}{2} m V^{2}=e V\)

[ \(V\) is the potential difference]

\(v = \sqrt {\frac{{2eV}}{{{m_d}}}} \)

But \(m_{d}=2 m\)

Therefore, \(v=\sqrt{\frac{2 e V}{2 m}}=\sqrt{\frac{e V}{m}}\)

Radius of path, \(R=\frac{m v}{e B}\)

Substituting value of \('V'\) we get

\(R = \frac{{2m\sqrt {\frac{{ev}}{m}} }}{{eB}}\)

\(\frac{R}{2}=\frac{m \sqrt{\frac{e v}{m}}}{e B}\)  .... \((i)\)

For proton :

\(\frac{1}{2} m V^{2}=e V\)

\(v=\sqrt{\frac{2 e V}{m}}\)

Radius of path,  \(R' = \frac{{mV}}{{eB}} = \frac{{m\sqrt {\frac{{2eV}}{m}} }}{{eB}}\)

\(R = \sqrt 2  \times \frac{R}{2}\)  [From eq. \((i)\) ]

\(R' = \frac{R}{{\sqrt 2 }}\)