Question

A certain commercial mass spectrometer is used to separate uranium ions of mass 3.92

Answer 1

a)

KE=(1/2)mv^2 =q*V

(1/2)*(3.92*10^-25)*v^2 =(3.2*10^-19)*(129*10^3)

v=4.59*10^5 m/s

Centripetal force =Magnetic force

mv^2/r=qvB

=>B=mv/qr =(3.92*10^-25)*(4.59*10^5)/(3.2*10^-19)*(1.24)

B=0.4534 T

b)

The number of ions/sec in 0.84 mg material is

0.84*10^-3 =(ions/sec)*3600*3.92*10^-25

Ions/sec =5.95*10^17

now current is

I=(5.95*10^17)*(3.2*10^-19)

I=0.19 A

c)

U=qV =3.2*10^-19*129*10^3 =4.128*10^-14 J/ion

Energy

E=(5.95*10^17)*(1.4*3600)*(4.128*10^-14)

E=1.238*10^8 J

Answer 2

I have solved this question earlier with different figures. Please workout using yours figures. If you need any further help just PM me. If I have helped you please rate me 5 stars first (before you rate anyone else).

A certain commercial mass spectrometer (Fig. 28-12) is used to separate uranium ions of mass 3.92 x 10-25 kg and charge 3.20 x 10-19 C from related species. The ions are accelerated through a potential difference of 109 kV and then pass into a uniform magnetic field, where they are bent in a path of radius 1.31 m. After traveling through 180