91Ó°ÊÓ

• $r=53.0cm=0.53m$
  
• $f=12.0MHz=12.0×{10}^{6}Hz$
  
• $B=1.57T$
  

We can find the relation between the frequency and the magnetic field using the relation$\mathit{q}\mathit{v}\mathit{B}\mathbf{=}\frac{\mathbf{m}{\mathbf{v}}^{\mathbf{2}}}{\mathbf{r}}$From this; we can solve for the B. We can find the equation for vin terms of f. We substitute the value for vin the formula of kinetic energy to get the kinetic energy of the proton.

We use the same relation, i.e.,$\mathit{q}\mathit{v}\mathit{B}\mathbf{=}\frac{\mathbf{m}{\mathbf{v}}^{\mathbf{2}}}{\mathbf{r}}$This time, we solve for f. We can find the equation for vin terms of f. We substitute the value for vin the formula of kinetic energy to get the kinetic energy of the proton emerging from the cyclotron.

Formula:

$\begin{array}{rcl}qvB& =& \frac{m{v}^2}{r}\\ K& =& \frac{1}{2}m{v}^2\\ \mathrm{Ó\neg }& =& v/r\\ \mathrm{Ó\neg }& =& 2\mathrm{Ï€}f\end{array}$

We know that:

$\begin{array}{rcl}qvB& =& \frac{m{v}^2}{r}\\ qB& =& \frac{mv}{r}\\ \frac{qB}{m}& =& \frac{v}{r}\\ \mathrm{Ó\neg }& =& \frac{qB}{m}\end{array}$

We know that:

$\begin{array}{rcl}\mathrm{Ó\neg }& =& 2\mathrm{Ï€}f\\ 2\mathrm{Ï€}f& =& \frac{qB}{m}\end{array}$

The magnitude of the magnetic field can be calculated as:

$\begin{array}{rcl}B& =& \frac{2\mathrm{Ï€´Ú³¾}}{q}\\ & =& \frac{2\mathrm{Ï€}×12.0×{10}^6×1.67×{10}^{-27}}{1.6×{10}^{-19}}\\ & =& 0.787T\end{array}$

The magnitude of the magnetic field to achieve resonance will be 0.787 T.

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

We know that

role="math" localid="1662880233360" $\begin{array}{rcl}\frac{v}{r}& =& omega\\ \frac{v}{r}& =& 2\mathrm{Ï€´Ú}\\ \mathrm{v}& =& 2\mathrm{Ï€°ù´Ú}\\ \mathrm{K}& =& \frac{1}{2}×\mathrm{m}×{\left(2\mathrm{Ï€°ù´Ú}\right)}^2\\ & =& \frac{1}{2}×1.67×{10}^{-27}×{\left(2\mathrm{Ï€}×0.53×12×{10}^6\right)}^2\\ & =& 1.33×{10}^{-12}\mathrm{J}\\ & =& 1.33×{10}^{-12}\mathrm{J}×\frac{\left(6.24×{10}^{18}\right)\mathrm{eV}}{1\mathrm{J}}\\ & =& 8.34×{10}^6\mathrm{eV}\end{array}$

The kinetic energy of the proton will be$\begin{array}{rcl}& & 8.34×{10}^6\mathrm{eV}\end{array}$

The frequency can be calculated as:

$\begin{array}{rcl}f& =& \frac{qB}{2\mathrm{Ï€³¾}}\\ & =& \frac{1.6×{10}^{-19}×1.57}{2\mathrm{Ï€}×1.67×{10}^{-27}}\\ & & \\ & =& 2.39×{10}^7\mathrm{Hz}\end{array}$

$\begin{array}{rcl}& & 2.39×{10}^7\mathrm{Hz}\end{array}$is the oscillator frequency to achieve resonance.

The expression for kinetic energy:

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

We know that

$$\begin{gathered} \frac{v}{r}=omega \\ \frac{v}{r}=2\mathrm{Ï€´Ú} \end{gathered}$$

The expression for velocity can be written as:

$\begin{array}{rcl}v& =& 2\mathrm{Ï€°ù´Ú}\\ \mathrm{K}& =& \frac{1}{2}×\mathrm{m}×{\left(2\mathrm{Ï€°ù´Ú}\right)}^2\\ & =& \frac{1}{2}×1.67×{10}^{-27}×{\left(2\mathrm{Ï€}×0.53×2.39×{10}^7\right)}^2\\ & =& 5.3×{10}^{-12}\mathrm{J}\\ & =& 5.3×{10}^{-12}\mathrm{J}×\frac{\left(6.24×{10}^{-18}\right)\mathrm{eV}}{1\mathrm{J}}\\ & =& 3.32×{10}^7\mathrm{eV}\end{array}$

The kinetic energy of the emerging proton will be$\begin{array}{rcl}& & 3.32×{10}^7\mathrm{eV}\end{array}$