An accelerator produces alphas using an electric potential difference V_0 from 1

Question

An accelerator produces alphas using an electric potential difference V_0 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 V_0 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. (Note: 1 u = 1.66 times 10^-27 kg)

Explanation / Answer

7) a) charge of alpha q = 2e

Vo = 2.5 MV

From conservation of energy

KE = qVo = 2*2.5 MeV = 5 MeV

b) r = 45 cm , mass m = 4*1.66*10^-27 kg

From conservation of energy KE =(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

magnetic force = centripetal force

qvB = mv^2/r

qB = mv/r

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

B =0.64 T

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