As charge on both proton and deuteron is same i.e. ' e ' Energy acquired by both, E=eV For Deuteron. Kinetic energy, 21mV2=eV [ V is the potential difference] v=md2eV But md=2m Therefore, v=2m2eV=meV Radius of path, R=eBmv Substituting value of ' v ' we get R=eB2mmev2R=eBmmev For proton : 21mV2=eVV=m2eV Radius of path, R′=eBmV=eBmm2eVR′=2×2R [From eq. (i)] R′=2R