What is the speed of an electron whose kinetic energy is 1.66MeV?

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a small but measurable current of 1.2x10^-10A exists in a copper wire whose diameter is 2.5mm. the number of charge carriers per unit volume is 8.49x10^28m^-3. Assuming the current is uniform, calculate the current density and electron drift speed
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Mass of electron=9.109*10^-31 kg; 1MeV=1.6022*10^-13 joule.

Kinetic energy=½mv^2 where m=mass of electron in kg, energy is in joules, v=metres/sec.

1.66*1.6022*10^-13=9.109*10^-31v^2/2.

2.9198*10^17=v^2/2; v=√5.84*10^17=7.642*10^8 m/s. However, the speed of light is 3*10^8 m/s and the speed of an electron cannot exceed the speed of light. Therefore, classical Newtonian physics cannot give the answer for energies of such magnitude.

At relativistic speeds, kinetic energy=mc^2-m0c^2 where m0=rest mass of the electron, which has an energy equivalent of 0.511MeV. So 1.66-0.51=1.15MeV=mc^2, where m is the relativistic mass of the electron. Speed of light, c=3*10^8 m/s. 

1.66MeV=2.66*10^-13 joule.

At high speeds the mass of a particle increases according to the Lorentz factor, so that more energy is required to increase its speed. Lorentz factor, L, is √(1-v^2/c^2) and energy=m0c^2(1/L-1)=2.66*10^-13.

1/L=1+(2.66*10^-13/(9.109*10^-31*9*10^16))=4.245; L=0.3082.

L^2=0.095=1-v^2/c^2; v^2/c^2=0.905, v/c=0.9513, v=0.9513c=2.85*10^8 m/s.

 

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