Answer:
(a).[tex]v_{avg} = 6.8*10^{4}m/s,[/tex] is 0.02% the speed of light.
(b).[tex]v_{avg} = 1.2*10^5m/s[/tex], is 0.04% the speed of light.
(c). [tex]v_{avg} =2.1*10^5m/s[/tex] is 0.07% the speed of light.
Explanation:
The average kinetic energy is related to the thermal energy of the electrons in a conductor by the the relation
[tex]K.E_{avg} = \dfrac{1}{2}mv_{avg}^2 = \dfrac{3}{2}kT[/tex],
where [tex]m = 9.1*10^{-31}kg[/tex] is the mass of the electrons, [tex]v_{avg}[/tex] is their average velocity, [tex]T[/tex] is the temperature of the conductor, and [tex]k = 1.38*10^{-23}m^2kg \:s^{-2}\:K^{-1}[/tex] is the Boltzmann constant.
The equation, when solved for [tex]v_{avg}[/tex], gives
[tex]v_{avg} = \sqrt{\dfrac{3kT}{m} }[/tex]
(a),
For [tex]T = 100K[/tex], the thermal (average) velocity [tex]v_{avg}[/tex] is
[tex]v_{avg} = \sqrt{\dfrac{3(1.38*10^{-23})(100K)}{9.1*10^{-31}kg} }[/tex]
[tex]\boxed{v_{avg} = 6.8*10^{4}m/s}[/tex]
which when compared to the speed of light is
[tex]\dfrac{6.8*10^4}{3*10^8} *100\% = (0.02\%)c[/tex]
0.02% the speed of light.
(b).
Similarly, for [tex]T =300K[/tex]
[tex]v_{avg} = \sqrt{\dfrac{3(1.38*10^{-23})(300K)}{9.1*10^{-31}kg} }[/tex]
[tex]\boxed{v_{avg} = 1.2*10^5m/s}[/tex]
which is
[tex]\dfrac{1,2*10^5m/s}{3*10^8m/s} *100\%= (0.04\%)c[/tex]
0.04% the speed of light.
(c).
Finally, for [tex]T =1000K[/tex]
[tex]v_{avg} = \sqrt{\dfrac{3(1.38*10^{-23})(1000k)}{9.1*10^{-31}kg} }[/tex]
[tex]\boxed{v_{avg} =2.1*10^5m/s}[/tex]
which is
[tex]\dfrac{2.1*10^5m/s}{3*10^8m/s} *100\% = (0.07\%)c[/tex]
0.07% the speed of light.
We see that the average electron velocities we obtain are always less than 1% the speed of light, which means relativistic effects are negligible, for they are apparent at about 25% the speed of light.
