A proton and a deutron both having the same kinetic energy, enter perpendicularly into a uniform magnetic field $B$. For motion of proton and deutron on circular path of radius ${R_p}$ and ${R_d}$ respectively, the correct statement is
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(a) $\frac{{m{v^2}}}{R} = qvB$. For proton ${R_p} = \frac{{mv}}{{qB}} = \frac{{\sqrt {2{m_p}E} }}{{qB}}$
and for deutron ${R_d} = \frac{{\sqrt {2{m_d}E} }}{{qB}}$
$ \Rightarrow \,\;\frac{{{R_d}}}{{{R_p}}} = \sqrt {\frac{{{m_d}}}{{{m_p}}}} = \sqrt 2 \Rightarrow {R_d} = \sqrt 2 {R_p}$
and for deutron ${R_d} = \frac{{\sqrt {2{m_d}E} }}{{qB}}$
$ \Rightarrow \,\;\frac{{{R_d}}}{{{R_p}}} = \sqrt {\frac{{{m_d}}}{{{m_p}}}} = \sqrt 2 \Rightarrow {R_d} = \sqrt 2 {R_p}$
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