The mass of a stationary electron is 9.11 times 10^-31 kg. (a)Using Einstein\'s
ID: 1997058 • Letter: T
Question
The mass of a stationary electron is 9.11 times 10^-31 kg. (a)Using Einstein's formula E = mc^2 find the total energy of an electron that is stationary - called the rest energy E_0 - of the electron. (1) (b)Find the mass of the electron when it is traveling at 3.0 times 10^7 m/s. (1) (c) Using the same formula as in (a), find the total energy E_s of the electron when it is moving at the speed given in (b). (1) (d) Now subtract the total energy of the electron at rest from the total energy of the electron in motion, i.e. find E_s - E_0. This is the kinetic energy of the electron. (1) (e) The electron does not obey classical mechanics at high speeds. But suppose we were to use the classical formula for kinetic energy 1/2mv^2 where m = 9.11 times 10^-31 kg and upsilon = 3.0 times 10^7 m/s. What would be the kinetic energy of the electron if it obeyed classical physics? (1) (f) Find the percentage difference between the two values of kinetic energy that you obtained in (d) and (e). Write your answer to 2 significant figures. (So if your answer is 15.7 % you would write 16 %. If your answer is 0.0141 % you would write 0.014 %.) (2)Explanation / Answer
m0 = 9.11 *10^-31 Kg
a) rest energy , Eo
rest energy , Eo = m0 * c^2
rest energy , Eo = 9.11 *10^-31 * (3 *10^8)^2
rest energy , Eo = 8.2 *10^-14 J
b)
for the mass of electron at 3 *10^7 m/s
mass of electron , m = m0/sqrt(1 - (v/c)^2)
m = 9.11 *10^-31/sqrt(1 - (3 *10^7/(3 *10^8))^2)
m = 9.16 *10^-31 Kg
c)
at the given speed ,
total energy = m * c^2
total energy = 9.16 *10^-31 * (3 *10^8)^2
total energy = 8.244 *10^-14 J
d) kinetic energy of electron = Ea - E0
kinetic energy of electron = 8.244 *10^-14 - 8.2 *10^-14
kinetic energy of electron = 4.4 *10^-16 J
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