Question

Exercise 27.20

Cyclotrons are widely used in nuclear medicine for producing short-lived radioactive isotopes. These cyclotrons typically accelerate H− (the hydride ion, which has one proton and two electrons) to an energy of 5MeV to 20MeV. This ion has a mass very close to that of a proton because the electron mass is negligible−about 1/2000 of the protons mass. A typical magnetic field in such cyclotrons is 1.7 T .

Part A

What is the speed of a 5.0-MeV H−?

Express your answer with the appropriate units.

Part B

If the H− has energy 5.0MeV and B= 1.7 T , what is the radius of this ions circular orbit?

Express your answer with the appropriate units.

Answer 1

Concepts and reason

The concepts used to solve this problem are energy conservation and the force conservation in a magnetic field for a circular motion.

Initially, the velocity can be calculated by using the formula from the energy conservation. Later the forces in a magnetic field for a circular motion have to equate to calculate the expression for the radius of the circular path. Finally, the radius of the circular path can be calculated by substituting the given numerical values in the problem.

Fundamentals

From energy conservation, the velocity of the Hydrogen is,

$\begin{array}{c}\\E = \frac{1}{2}m{v^2}\\\\v = \sqrt {\frac{{2E}}{m}} \\\end{array}$

Here, $E$is the energy of the hydrogen, $v$is the speed of the hydrogen, and $m$is the mass of proton.

At equilibrium, the forces in a magnetic field is,

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

Here, $B$is the magnetic field, $q$is the charge of proton and $r$is the radius of the circular orbit.

(A)

The expression for the speed is,

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

Here, $E$is the energy of the hydrogen, $v$is the speed of the hydrogen, and $m$is the mass of proton.

Substitute $5\;{\rm{MeV}}$for$E$, $1.67 \times {10^{ - 27}}{\rm{kg}}$for$m$.

$\begin{array}{c}\\v = \sqrt {\frac{{2 \times \left( {5\;{\rm{MeV}}} \right)\left( {\frac{{1 \times {{10}^6}{\rm{eV}}}}{{1\;{\rm{MeV}}}}} \right)\left( {\frac{{1.6 \times {{10}^{ - 19}}{\rm{J}}}}{{1\;{\rm{eV}}}}} \right)}}{{1.67 \times {{10}^{ - 27}}\,{\rm{kg}}}}} \\\\ = 3.09 \times {10^7}{\rm{m/s}}\\\end{array}$

(B)

For a charged object to move in a circular path under uniform magnetic field, the magnetic force and centripetal force acting on the charged particle must be balanced.

The expression for the radius of the circular path is,

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

Substitute $1.67 \times {10^{ - 27}}{\rm{kg}}$for$m$, $3.09 \times {10^7}{\rm{m/s}}$for$v$, $1.6 \times {10^{ - 19}}{\rm{C}}$for $q$ and $1.7\;{\rm{T}}$for$B$

$\begin{array}{c}\\r = \frac{{\left( {1.67 \times {{10}^{ - 27}}{\rm{ kg}}} \right) \times \left( {3.09 \times {{10}^7}{\rm{ m/s}}} \right)}}{{\left( {1.6 \times {{10}^{ - 19}}{\rm{ C}}} \right) \times \left( {1.7{\rm{ T}}} \right)}}\\\\\;\; = 0.189\;{\rm{m}}\\\end{array}$

Ans: Part A

The speed of the hydrogen is$3.09 \times {10^7}\,{\rm{m/s}}$.

Part B

The radius of the circular orbit is$0.189\;{\rm{m}}$