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.
 

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
What are you looking for?