Answer:
So voltage drop will be 16.7125 volt
Explanation:
We have distance traveled by proton = 6.40 cm = 0.064 m
Time = [tex]T=1.13\times 10^{-6}sec[/tex]
So velocity [tex]v=\frac{distance}{time }=\frac{0.064}{1.13\times 10^{-6}}=0.0566\times 10^6m/sec[/tex]
Kinetic energy of proton is given by [tex]E=\frac{1}{2}mv^2=\frac{1}{2}\times 1.67\times 10^{-27}\times (0.0566\times 10^6)^2=2.674\times 10^{-18}j[/tex]
And potential energy of proton is given by [tex]E=eV=1.6\times 10^{-19}\times V[/tex]
Both potential energy and kinetic energy will be same
So [tex]1.6\times 10^{-19}\times V=2.674\times 10^{-18}[/tex]
[tex]V=16.7125Volt[/tex]
