Question

In a nuclear experiment a proton with kinetic energy 4.0 MeV moves in a circular path in a uniform magnetic field. If the magnetic field is B = 1.6 T what is the radius of the orbit?

What energy must an alpha particle (q = +2e, m = 4.0 u) and a deuteron (q = +e, m = 2.0 u) have if they are to circulate in the same orbit? Energy of the alpha? (MeV)

Energy of the deuteron?

Answer 1

KE = mv^2 /2

4 x 10^6 x 1.6 x 10^-19 J = 1.673 x 10^-27 kg x v^2 /2

v = 2.77 x 10^7 m/s

using mv^2 / R = qvB

R = mv / qB

for proton : R = ( 1.673x 10^-27 x 2.77 x 10^7 ) / (1.6 x10^-19 x 1.6)

R = 0.181 m

for alpha :

R = mv / qB

0.181 = (4 x 1.673 x 10^-27 x v ) / (2 x 1.6 x 10^-19 x 1.6)

v = 1.38 x 107 m/s

KE = 4 x 1.673 x 10^-27 x (1.38 x 10^7)^2 /2 = 1.28 x 10^-12 J

KE ( in eV) = 1.28 x 10^-12 / 1.6 x10^-19 = 8 x 10^6 eV = 8 MeV

for deutron:

0.181 = (2 x 1.673 x 10^-27 x v ) / (1.6 x 10^-19 x 1.6)

v = 1.38 x 107 m/s

KE = 2 x 1.673 x 10^-27 x (1.38 x 10^7)^2 /2 = 0.64 x 10^-12 J

KE ( in eV) = 0.64 x 10^-12 / 1.6 x10^-19 = 4 x 10^6 eV = 4 MeV