Question

a) A mass spectrometer is designed to separate molecules out by charge/mass ratios. A test molecule weighing 3.32_E-25 kg is first ionized to a +2 charge and sent through the mass spectrometer. If it traces out a semi-circle with radius of 10 cm in a magnetic field of 1.805 T, what is its velocity and what electric field is necessary to select this velocity?

b) If a second molecule weighing 4.15_E-25 kg is ionized to a -3 charge and sent through the mass spectrometer, what will be the separation between it and the first molecule?

Answer 1

given
m = 3.32*10^-25 kg
q = +2*e

= +2*1.6*10^-19 C

= 3.2*10^-19 C
r = 10 cm = 0.1 m
B = 1.805 T

v = ?

we know, r = m*v/(B*q)

==> v = B*q*r/m

= 1.805*3.2*10^-19*0.1/(3.32*10^-25)

= 1.74*10^5 m/s <<<<<<<<<-------------Answer

for velocity selction, Fe = FB

q*E = q*v*B

E = v*B

= 1.74*10^5*1.805

= 3.14*10^5 N/C <<<<<<<<<-------------Answer

b) we know, r = m*v/(B*q)

here v and B are same for both Ions.

r2/r1 = (m2/m1)/(q2/q1)

r2 = r1*(m2/m1)/(q2/q1)

= 0.1*(4.15/3.32)/(3/2)

= 0.083 m or 8.3 cm

so, separation between the two charges = 2*(10 - 8.3) (after completing half circle)

= 3.4 cm <<<<<<<<<-------------Answer