question 1: question 1: A doubly ionized molecule is studied in a mass spectrome

Question

question 1:

question 1: A doubly ionized molecule is studied in a mass spectrometer, where it is accelerated through a potential difference of 410 V and found to travel in a circular path of radius 7.70 cm in a 0.460-T magnetic field. What is the molecule’s mass?please give me the answer in kg.

question 2:

question 2: At what speed will a particle of mass 9.7×109 kg and charge 27 nC move in a circular path of radius 5.2 cm in a uniform 2.2-T magnetic field? please give me the answer in m/s.

question 3: A particle of mass 7.8×106 kg and charge 61 C is moving at a speed of 43 m/s perpendicularly to a uniform 6.1-T magnetic field. What is the radius of the particle’s circular path?

question 1: A doubly ionized molecule is studied in a mass spectrometer, where it is accelerated through a potential difference of 410 V and found to travel in a circular path of radius 7.70 cm in a 0.460-T magnetic field. What is the molecule’s mass?please give me the answer in kg.

question 2:

question 2: At what speed will a particle of mass 9.7×109 kg and charge 27 nC move in a circular path of radius 5.2 cm in a uniform 2.2-T magnetic field? please give me the answer in m/s.

question 3:

question 3: A particle of mass 7.8×106 kg and charge 61 C is moving at a speed of 43 m/s perpendicularly to a uniform 6.1-T magnetic field. What is the radius of the particle’s circular path?

question 4: The electric field in a velocity selector is generated by two parallel plates separated by 5.00 cm with a potential difference of 90.0 V. If the magnetic field is 1.300e-1 T, at what speed will particles pass through undeflected? Recording plate AV

Explanation / Answer

1. given the molecule is doublty ionised, so charge on particle = 2e ( where e is charge on ele)cron)
potential difference, V = 410 V
radius of path, r = -0.077 m
magnetic field, B = 0.46 T
let the mass of the molecule be m
then
KE after getting accelerated by potential V, KE = 2eV = 0.5mv^2 ( where v is the velocity of the molecule)
also, from balancing centripital and magnetic force
2evB = mv^2/r
v = 2eBr/m = sqroot(4eV/m)
e*B^2*r^2/V = m = 4.89*10^-25 kg
2. m = 9.7*10^-9 kg
q = -27*10^-9 C
r = 0.052 m
B = 2.2 T
v = ?
mv/r = qB
v = qBr/m = 0.31843 m/s
3. m = 7.8*10^-6 kg
q = 61 micro C
v = 43 m/s
B = 6.1 T
mv/r = qB
r = mv/qB = 0.9013 m

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