Question

A cyclotron accelerates a proton to a final speed of 3×10^7ms^-1,which is initially at rest.The work done on the proton in mega electron volts,by the electrical field of the cyclotron is

Answer 1

work energy Theorem...

work = change in kinetic energy

$work = \frac{1}{2} m {v}^{2} \\ = \frac{1}{2} \times 1.6 \times {10}^{ - 27} \times {(3 \times {10}^{7} )}^{2} \\ joules$
=
$= \frac{1}{2} \times 1.6 \times {10}^{ - 27} \times {(3 \times {10}^{7}) }^{2} \div (1.6 \times {10}^{ - 19} ) \: ev$
$= 4.5 \times {10}^{6} ev$
= 4.5 MeV