# Protons in Cyclotron (Nuclear Physics)

What is the total distance traveled by the protons and by the $\alpha$ particles in a cyclotron that has the effective acceleration voltage between the “D”-shaped electrodes (“d-ees”)  $U =25 kV$. The frequency of the voltage generator is $\nu= 10^{7} Hz$ and the maximum radius of the particles trajectory is $R =0.5 m$. The space between the “D”-shaped electrodes is negligible and the number of rotations of particles until they reach the desired energy is considered very big. Take $m_p= 1 amu$, $m_{\alpha} =4 amu$, $e =1.6*10^{-19} C$.

If $V_n$ is the speed of a particle after passing the $n$-th time through the acceleration space between the d-ees (one “d-ee” is half of a complete circle) then one can write the total distance traveled as:

$L = \sum_{n=1}^{N}V_n*(T/2) =1/(2\nu)*\sum_{n=1}^{N}V_n$

From the energy equality

$\frac{m V_n^2}{2} =n q U$

one can write the speed $V_n$ as

$V_n = \sqrt{(2nqU)/m}$

and therefore the distance traveled is

$L = (1/\nu)\sqrt {\frac{q u}{2m}}*\sum_{n=1}^N \sqrt n$

Since the number of consecutive accelerations (passing between d-ees) $N$ is very big

$N =E_{max}/(q U)$

one can write

$\sum_{n=1}^{N}\sqrt n \rightarrow \int_{0}^{N}\sqrt n*d n=(2/3)*N^{(3/2)}=\frac {2}{3}\left ( \frac{2\pi^2 m \nu^2 R^2}{q U} \right )^{3/2}$

and therefore the total distance traveled by the particles until they reach the maximum energy is

$L =(4\pi^3 m \nu^2 R^3)/(3qU)$

With the numerical values given one has:

$L_p =214.46 m$

$L_{\alpha} =2L_p =428.92 m$