Question

A proton is accelerating in a cyclotron where the applied magnetic field is 2T . If the potential gap is effectively 100kV then how much revolutions the proton has to make between the dees to acquire a kinetic energy of 20MeV?

Answer 1

The kinetic energy $E$ of the particle can be calculated with the initial speed 0 when it leaves the particle source and the number $n$ of passages through the electric field between the "dees",

We have $nqV=E\\ n=\frac{E}{qV} \\ n=\frac{100*1000}{20*1.602*10^{-19}*10^6} \\ n=3.12*10^{16}$

The number of revolutions the proton has to make between the dees is

$\frac{3.12*10^{16}}{2} =1.56*10^{16}$

Answer 2

Answer:. V=100kV

Kinetic energy = 2MeV

= 20× 10^6×1.6×10^-19

No. Of turns =KE/2qV

=20×10^6×1.6×10^-19/2×1.6×10^-19×100×10^3

=100

Step-by-step explanation: