91Ó°ÊÓ

• $r_{\text{cyclotron}}=53cm=0.53m$
  

• Operating frequency$f=12MHz$
• Accelerating potential between Dees $80keV$.

We are given the final energy of the deuteron. We know that it is accelerated twice in each cycle, and the energy it gains when it is accelerated. From this, we can find the number of revolutions made by the deuteron. From this number, we can find the velocity of the deuteron and the radius of the path. Using the radius of the path and the number of revolutions made, we can find the total distance traveled.

Formula:

$n=\frac{\text{Finalenergy}}{\text{Energygainedperrevolution}}$

$r=\frac{mv}{qB}$

$K=\frac{1}{2}m{v}^2$

We know that:

$n=\frac{\text{Finalenergy}}{\text{Energygainedperrevolution}}$

The deuteron accelerates twice in each cycle; each time, it receives the energy of $80×{10}^{3}eV$. We are given the final energy as $16.6×{10}^{6}eV$.

$$\begin{gathered} n=\frac{16.6×{10}^{6}eV}{2×\left(80×{10}^{3}eV\right)} \\ =103.75 \\ ≈104 \end{gathered}$$

We know that:

$K=\frac{1}{2}m{v}^2$

$v=\sqrt{\frac{2K}{m}}$

We know that:

$r=\frac{mv}{qB}$

$r=\frac{m}{qB}×\left(\sqrt{\frac{2K}{m}}\right)$

$r=\frac{\sqrt{2Km}}{qB}$

$$\begin{gathered} K=8.3×{10}^6×1.6×{10}^{-19}J \\ =1.328×{10}^{-12}J \end{gathered}$$

$$\begin{gathered} r=\frac{\sqrt{2×1.328×{10}^{-12}×3.34×{10}^{-27}}}{1.6×{10}^{-19}×1.57} \\ =0.37494\text{m} \end{gathered}$$

The total distance traveled will be:

$$\begin{gathered} \text{Total Distance}=n×2\mathrm{π}r \\ =103.75×2\mathrm{π}×0.375494 \\ =244.416 \\ =2.4×{10}^2\text{m} \end{gathered}$$

The total path length traveled by deuteron will be $2.4×{10}^{2}m$.