Question

1) An electron has a kinetic energy that is 50% larger than its classical kinetic energy. Electron mass is 0.511 MeV/c^2.

a. What is the speed of the electron expressed in the unit of speed of light c?

b. What is the total energy of the electron expressed in the unit of MeV?

c. What is the kinetic energy of the electron expressed in the unit of MeV?

Answer 1

a )

the momentum in classical is P_cl = P_o = m_o v

relativistic momentum P = m v

m = m_o / ( 1 -² )^1/2

= 1 / ( 1 -² )^1/2

P = m_o v / ( 1 -² )^1/2

⁼ P_cl

given

momentum is 50 % larger than P_cl

P = P_cl + 0.5 P_cl

= 3 P_cl / 2

P / P_cl = = 3/2

1/ ² = 4/9

(1 -² ) = 4/ 9

= 0.74

v = C

= 0.74 C

c )

using KE = ( P² C² + m_o² C⁴ )^1/2

= m_oC² ( ²² + 1 )^1/2

= 0.511 x ( 5/4 +1 )^1/2

KE = 0.7665 MeV

b )

T = KE + rest mass energy

KE = T + m_o C²

T = KE - m_o C²

=0.5 m_o C²

= 0.5 x 0.511

T = 0.255 MeV