Question

(i) Compute the value of 𝛾 for a particle traveling at half the speed of light. Give your answer

to three significant figures.

(ii) Determine the energy required to give an electron a speed of 0.90 that of light, starting from

rest.

Answer 1

(i) We find the value for gamma with

$γ=\frac{1}{\sqrt{1−(v/c)²}} \\ \gamma =\frac{1}{\sqrt{1−(0.500)²}}=\frac{1}{\sqrt{0.750}} \\ \therefore \gamma =\frac{1}{0.866}=1.15$

(ii) The kinetic energy required is equal to

$KE = (\gamma m - m)c²=(\gamma - 1)mc² \\ \therefore KE= mc² \Big(\frac{1}{\sqrt{1−(v/c)²}}-1\Big) \\ \text{We substitute and find KE} \\ KE = (\frac{1}{\sqrt{1−(0.90)²}}-1)(9.11×10^{-31}{kg})(3×10⁸\frac{m}{s})² \\ \therefore KE = 1.06×10^{-13}J$

In conclusion, (i) 𝛾 = 1.15 when v/c = 0.50, while (ii) we need 1.06 × 10^-13 J of kinetic energy to take an electron from the rest up to v/c = 0.90.