Answer:
The electric field inside the wire will remain the same or constant, while the drift velocity will by a factor of four.
Explanation:
Electron mobility, μ =
where
= Drift velocity
E = Electric field
Given that the electric field strength = 1.48 V/m,
Therefore since the electric potential depends on the length of the wire and the attached potential difference, then when the electron mobility is increased 4 times the Electric field E will be the same but the drift velocity will increase four times. That is
4·μ = 
Answer:
50 KeV
Explanation:
50 KeV will be the most likely initial interaction with tissue.
Because Higher the energy lesser will the interaction with tissues, Because it will penetrate through tissue more easily in a very lesser time Whereas the photon with least energy will take more time to penetrate through the tissue, hence, higher interaction with tissue.