For the reference frame of the electron, calculate the distance travelled by the detector.

For the reference frame of the laboratory, calculate the time taken for the electron to reach the detector after its emission from the nucleus.

For the reference frame of the electron, calculate the time between its emission at the nucleus and its detection.

Outline why the answer to (c) represents a proper time interval.

γ=4.503

\( \ll \frac{{0.800}}{{4.50}} =  \gg 0.178{\rm{m}}\)

\({\rm{time}} = \frac{{0.800}}{{2.94 \times {{10}^8}}}\)

2.74 ns

\(\frac{{2.74}}{{4.5}}\) OR \(\frac{{0.178}}{{2.94 \times {{10}^8}}}\)

0.608 ns

it is measured in the frame of reference in which both events occur at the same position
OR
it is the shortest time interval possible