Solved 1 Answer
See More Answers for FREE
Enhance your learning with StudyX
Receive support from our dedicated community users and experts
See up to 20 answers per week for free
Experience reliable customer service
Using Energy conservation in between both acceleratingplates:KEi + PEi = KEf + PEfKEi = 0PEi - PEf = change in potential energy = q*DeltaVKEf = (1/2)*m*v^2v = velocity of ion when it enters in magnetic field(1/2)*m*v^2 = q*DeltaVDeltaV =m*v^2/(2*q)Now when ion enters into magnetic field using force balance onitFc = FmCentripetal force on ion = magnetic forcem*v^2/R = q*v*Bv = q*B*R/mSo, DeltaV =m*(q*B*R/m)^2/(2*q)DeltaV =q*B^2*R^2/(2*m)Given that for oxygenmass of 16O = 15.992 umass of O2+ = 2*massof 16O = 2*15.992*1.6605*10^-27 kgq = charge on oxygen O2+ = 1e =1*1.6022*10^-19B = magnetic field = 200.00 mT = 200.00*10^-3 TR = radius of path = half of distance between entrance and exitholes = 8.0000 cm/2 = 4.0000 cm = 4.0000*10^-2 mNow Using these values:DeltaV =(1*1.6022*10^-19*(200.00*10^-3)^2*(4.0000*10^-2)^2)/(2*2*15.992*1.6605*10^-27)DeltaVO2+ = 96.537VPart B. Given that for oxygenmass of 14N = 14.003 umass of N2+ = 2*massof 14N = 2*14.003*1.6605*10^-27 kgq = charge on oxygen N2+ = 1e =1*1.6022*10^-19B = magnetic field = 200.00 mT = 200.00*10^-3 TR = radius of path = half of distance between entrance and exitholes = 8.0000 cm/2 = 4.0000 cm = 4.0000*10^-2 mNow Using these values:DeltaV =(1*1.6022*10^-19*(200.00*10^-3)^2*(4.0000*10^-2)^2)/(2*2*14.003*1.6605*10^-27)DeltaVN2+ = 110.25VPart C. for CO+mass of CO+ = mass of 12C + massof 16O = (12.000 + 15.992)*1.6605*10^-27 kgq = charge on oxygen N2+ = 1e =1*1.6022*10^-19B = magnetic field = 200.00 mT = 200.00*10^-3 TR = radius of path = half of distance between entrance and exitholes = 8.0000 cm/2 = 4.0000 cm = 4.0000*10^-2 mNow Using these values:DeltaV =(1*1.6022*10^-19*(200.00*10^-3)^2*(4.0000*10^-2)^2)/(2*(12.000 +15.992)*1.6605*10^-27)DeltaVN2+ = 110.30 VPlease Upvote.  ...