Question

A cyclotron is used to produce a beam of high-energy deuterons that then collide with a target to produce radioactive isotopes for a medical procedure. Deuterons are nuclei of deuterium, an isotope of hydrogen, consisting of one neutron and one proton, with total mass 3.34×10−27kg. The deuterons exit the cyclotron with a kinetic energy of 6.40 MeV .

A-What is the speed of the deuterons when they exit?

B-If the magnetic field inside the cyclotron is 1.25 T, what is the diameter of the deuterons' largest orbit, just before they exit?

C-If the beam current is 350 μA how many deuterons strike the target each second?

Answer 1

Concepts and reason

The concept required to solve this problem is kinetic energy, centripetal force, and current in case of cyclotron.

Initially, write an expression for the kinetic energy and solve for the speed. Later, calculate the centripetal force by using the relation between the centripetal force, radius, and speed and the magnetic force by using the magnetic force equation when a charge is moving in a magnetic field. Finally, calculate the current by using the relation between the current and charge and time.

Fundamentals

The expression for the kinetic energy is as follows:

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

Here, m is mass ad v is the speed.

The expression for the centripetal force is as follows;

${F_{\rm{C}}} = \frac{{m{v^2}}}{r}$

Here, r is the radius of the circular track.

The expression for the magnetic force is as follows:

${\vec F_{\rm{B}}} = q\left( {\vec v \times \vec B} \right)$

Here, q is the charge, $\vec v$ is the velocity vector, and $\vec B$ is the magnetic field vector.

The expression for the current is as follows:

$i = \frac{{nq}}{t}$

Here, n is the number of electrons and t is the time taken.

(A)

Rearrange the equation $T = \frac{1}{2}m{v^2}$ for v.

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

Now, substitute 6.40 MeV for T and $3.34 \times {10^{ - 27}}{\rm{ kg}}$ for m in the equation $v = \sqrt {\frac{{2T}}{m}}$ .

$\begin{array}{c}\\v = \sqrt {\frac{{2\left( {6.40{\rm{ MeV}}} \right)}}{{3.34 \times {{10}^{ - 27}}{\rm{ kg}}}}} \\\\ = \sqrt {\frac{{2\left( {6.40{\rm{ MeV}}} \right)\left( {\frac{{{{10}^6}{\rm{ eV}}}}{{1.0{\rm{ MeV}}}}} \right)}}{{3.34 \times {{10}^{ - 27}}{\rm{ kg}}}}} \\\\ = \sqrt {\frac{{2\left( {6.40{\rm{ MeV}}} \right)\left( {\frac{{{{10}^6}{\rm{ eV}}}}{{1.0{\rm{ MeV}}}}} \right)\left( {\frac{{1.6 \times {{10}^{ - 19}}{\rm{ J}}}}{{1.0{\rm{ eV}}}}} \right)}}{{3.34 \times {{10}^{ - 27}}{\rm{ kg}}}}} \\\\ = 2.48 \times {10^7}{\rm{ m/s}}\\\end{array}$

(B)

Equate the equation ${F_{\rm{C}}} = \frac{{m{v^2}}}{r}$ and ${F_{\rm{B}}} = qvB$ .

$\begin{array}{c}\\\frac{{m{v^2}}}{r} = qvB\\\\\frac{{mv}}{r} = qB\\\end{array}$

Substitute $\frac{d}{2}$ for r in the equation $\frac{{mv}}{r} = qB$ .

$\begin{array}{c}\\\frac{{mv}}{{\frac{d}{2}}} = qB\\\\\frac{{2mv}}{d} = qB\\\end{array}$

Rearrange the equation $\frac{{2mv}}{d} = qB$ for d.

$d = \frac{{2mv}}{{qB}}$

Now, substitute $3.34 \times {10^{ - 27}}{\rm{ kg}}$ for m, $2.48 \times {10^7}{\rm{ m/s}}$ for v, 1.25 T for B, and $1.6 \times {10^{ - 19}}{\rm{ C}}$ for q in the equation $d = \frac{{2mv}}{{qB}}$ .

$\begin{array}{c}\\d = \frac{{2\left( {3.34 \times {{10}^{ - 27}}{\rm{ kg}}} \right)\left( {2.48 \times {{10}^7}{\rm{ m/s}}} \right)}}{{\left( {1.6 \times {{10}^{ - 19}}{\rm{ C}}} \right)\left( {1.25{\rm{ T}}} \right)}}\\\\ = 0.828{\rm{ m}}\\\end{array}$

(C)

Rearrange the equation $i = \frac{{nq}}{t}$ for n.

$n = \frac{{it}}{q}$

Now, substitute $350{\rm{ }}\mu {\rm{A}}$ for I, 1.0 s for t, and $1.6 \times {10^{ - 19}}{\rm{ C}}$ for q in the equation $n = \frac{{it}}{q}$ .

$\begin{array}{c}\\n = \frac{{\left( {350{\rm{ }}\mu {\rm{A}}\left( {\frac{{{{10}^{ - 6}}\;{\rm{A}}}}{{1\;\mu {\rm{A}}}}} \right)} \right)\left( {1{\rm{ s}}} \right)}}{{1.6 \times {{10}^{ - 19}}{\rm{ C}}}}\\\\ = 2.2 \times {10^{15}}\\\end{array}$

Ans: Part A

The speed of the deuteron is $2.48 \times {10^7}{\rm{ m/s}}$ .

Part B

The diameter of the deutron’s largets orbit is 0.828 m.

Part C

The number of deutrons are equal to $2.2 \times {10^{15}}$ .