Question

When a high‑energy photon passes near a heavy nucleus, a process known as pair production can occur. As a result, an electron and a positron (the electron's antiparticle) are produced. In one such occurrence, a researcher notes that the electron and positron fly off in opposite directions after being produced, each traveling at speed 0.941c. The researcher records the time that it takes for the electron to travel from one position to another within the detector as 15.7 ns. How much time would it take for the electron to move between the same two positions as measured by an observer moving along with the positron?

Answer 1

Answer:

1.47*10^{-8}s

Explanation:

You first calculate the distance traveled by the electron:

$x=vt\\\\x=(0.941(3*10^8m/s))(15.7*10^{-9}s)=4.43m$

Next, you calculate the relative speed as measure by an observer in the positron, of the electron:

$u'=\frac{u+v}{1+\frac{uv}{c^2}}\\\\u'=\frac{0.941c+0.941c}{1+\frac{(0.941)^2c^2}{c^2}}\\\\u'=0.99c$

with this relative velocity you calculate the time:

$t=\frac{x}{u'}\\\\t=\frac{4.43m}{0.99c}=1.47*10^{-8}s$