QJEEM22Ph9118: JEE Main 2022 Physics - Class 12 JEEM22Ph9118 Solution: A proton, a deutron and an -particle with same kinetic energy…

$\therefore$ $r = {{mv} \over {qB}} = {{\sqrt {2m(KE)} } \over {qB}}$

$\Rightarrow {r_1}:{r_2}:{r_3} = {{\sqrt {{m_1}} } \over {{q_1}}}:{{\sqrt {{m_2}} } \over {{q_2}}}:{{\sqrt {{m_3}} } \over {{q_3}}}$

$= {{\sqrt 1 } \over 1}:{{\sqrt 2 } \over 1}:{{\sqrt 4 } \over 2}$

$= 1:\sqrt 2 :1$