no.41 Q41. Fe + ions are accelerated through a potential difference of 500 V and are injected normally into a homogeneous magnetic field B of strength 20-0 mT. Find the radius of the circular paths followed by the isotopes with mass number 57 and 58. Take the mass of an ion = A(1.6 × 10–27) kg where A is the mass number. Share with your friends Share 3 Sandeep K. Gupta answered this Dear Student, Given- Charge on Fe+= +e , V=500 V , M1=57 ×1.6×10-27 kg and M2=58 ×1.6×10-27 kg and B=20.0×10-3 TBecause accelerating potential is same hence both the isotopes will have same kinetic energyK=qV -----(1)Radius of circular orbit in terms of KE is given byr=mvqB=2mKqB=2m(qV)qB=1B2mVq -----(2)For M1=57 ×1.6×10-27 kg r1=1B2×57 ×1.6×10-27 kg ×Vq=120.0×10-32×57×1.6×10-27×5001.6×10-19=1.193×10-2m=1.193 cmFor M2=M2=58 ×1.6×10-27 kg r2=1B2×58 ×1.6×10-27 kg ×Vq=120.0×10-32×58×1.6×10-27×5001.6×10-19=1.204×10-2m=1.204 cm Regards. 1 View Full Answer