Question

It is not very difficult to accelerate an electron to a speed that is 99.5% of the speed of light, because it has such a very small mass. what is the ratio of the kinetic energy k to the rest energy mc2 in this case? in the definition of what we mean by kinetic energy (k = e - mc2), you must use the full relativistic formula for e, because v/c is not small compared to 1.

Answer 1

The kinetic energy (K) to rest energy (mc2) ratio for an electron traveling at 99.5% of the speed of light is approximately 150%. This is determined using the full relativistic formula, as classical mechanics significantly underestimates the required energy.

Step-by-step explanation:

When an electron is accelerated to a speed that is 99.5% of the speed of light, relativistic effects become significant. The rest mass energy of an electron is 0.511 MeV, and at this high velocity, the kinetic energy is no longer accurately described by classical physics, and the full relativistic energy formula must be utilized. The relativistic kinetic energy of the electron is significantly greater than the classical value. If the classical kinetic energy is denoted as KEclass and the relativistic kinetic energy is denoted as KErel, the ratio KErel/KEclass is approximately 12.4. This highlights the substantial energy needed to accelerate a mass close to the speed of light, much more than classical physics would predict.

The ratio of the kinetic energy (K) to the rest energy (mc2) can be calculated using the relativistic energy equation E2 = (pc)2 + (mc2)2, where E is the total energy and p is the momentum. For an electron at 99.5% of the speed of light, this ratio is approximately 150%, meaning the kinetic energy is about one and a half times the rest mass energy. Therefore, for an electron moving at such high velocities, enormously high energies are required as indicated by currents accelerators like SLAC.

Answer 2

m = m_o / √(1-V^2/c^2 ) = 0.995 c

K/(m_o c^2 )=((m_o/√(1-v^2/c^2 )) v^2)/(1/2 m_o c^2 )=(m_o/√(1-v^2/c^2 ) (0.995)^2 c^2)/(1/2 m_o c^2 )=(2 x (0.995)^2)/√(1-(0.995c)^2/c^2 )= 198.5

k/m_oc^2 = 198.5