Answer
3.7.
Work Step by Step
The work done is the particle’s change in kinetic energy. Apply equation 26–5b. At rest, the kinetic energy is 0.
$$W_1=(\gamma_{0.90}-1)mc^2 - 0$$
$$W_2=(\gamma_{0.99}-1)mc^2 - (\gamma_{0.90}-1)mc^2=(\gamma_{0.99}-\gamma_{0.90})mc^2$$
Make a calculation of the $\gamma$ factors.
$$\gamma_{0.90}=\frac{1}{\sqrt{1-0.90^2}}=2.29416$$
$$\gamma_{0.99}=\frac{1}{\sqrt{1-0.90^2}}=7.08881$$
Find the requested ratio.
$$\frac{W_2}{W_1}=\frac{(\gamma_{0.99}-\gamma_{0.90})mc^2}{(\gamma_{0.90}-1)mc^2}$$
$$=\frac{7.08881-2.29416}{2.29416-1}=3.7$$
