An accelerator produces alphas using an electric potential difference Vo from 1.

Question

An accelerator produces alphas using an electric potential difference Vo from 1.50 MV to 9.00 MV as shown below: What is the kinetic energy of the alphas as they leave the accelerator if the electric potential difference Vo on the accelerator is 2.50 MV and they initially start from rest? An analyzing magnet supplies a uniform magnetic field which is perpendicular to the path of the alphas. Calculate the magnitude of the magnetic field required to select a 4.00 MeV alpha for a radius of curvature of 45.0 cm.

Explanation / Answer

V =2.5 MV , charge of alpha particle q = 2*1.6*10^-19 C

mass of alpha particle m = 4*1.66*10^-27 kg

a) From conservation of energy

K = U = qV

K = 2*e*2.5*MV = 5 MeV

K = 5*10^6*1.6*10^-19 = 8*10^-13 J

b) r= 45 cm

K = (1/2)mv^2 = 4 MeV

0.5*4*1.66*10^-27*v^2 = 4*10^6*1.6*10^-19

v = 1.388*10^7 m/s

centripetal force = magnetic force

mv^2/r = qvB

mv/r = qB

(4*1.66*10^-27*1.388*10^7)/(0.45) = 2*1.6*10^-19*B

B =0.64 T

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