Distance travelled in a Cyclotron

[Uku]

Homework Statement

Known data:
$$v_{final}=3.99*10^{7}m/s$$
$$f_{osc}=12MHz$$
$$B=1.6T$$
$$R=0.53m$$
$$V_{acc}=80kV$$
$$q_{deut}=1.602*10^{-19}C$$
$$m_{deut}=3.344*10^{-27}kg$$
answer according to Halliday and Resnick x=240m

I got it all, and I have to find how much distance was traveled by the deuteron in the cyclotron, from starting of acceleration to the end of it.

The Attempt at a Solution

From the potential across the dees I get

$$v_{gained_per_turn}=\sqrt{\frac{2E}{m_{deut}}}=2768586.602m/s$$

Since I know the final speed I can find the amount of turns the deuteron was in the cyclotron

$$n=\frac{v_{final}}{v_{gained_per_turn}}=14.41turns$$

Since f is constant, so is T and I can find the amount of time spent in the cyclotron:

$$t_{spent}=n*T=n*\frac{1}{f_{osc}}=1.2*10^{-6}s$$

The acceleration must be constant, the accelerating voltage is not changing

$$a=\frac{v_{gained_per_turn}}{T}=3.32*10^{13}\frac{m}{s^{2}}$$

Now

$$x=\frac{1}{2}a*t^{2}=23.92m$$

The answer in the book is 240m, which differs from mine ten times, smells like a power error?
Or am I doing something wrong?

 

[beamie564]

you don't need to find the time and stuff..
Change in Kinetic energy, ΔK=2qΔV (ΔV is the difference in potential and factor 2 because the particle is accelerated twice in one cycle)
Final Kinetic energy is K_f=(qBR)²/2m (use B=2πf_oscm)
then no. of cycles, n=K_f/ΔK
Total distance covered = n×2πR_avg (where R_avg = R/√2)
You'll get around 244m