Question

A beam of singly ionized uranium atoms 235 U and 238 U is injected into the mass spectrometer shown in the figure. The ions all have the same velocities and charges. The 238 U ions follow the trajectory illustrated.

1)Will the 235mU ions strike the collecting plate above, below, or at the same location as the 238mU ions?

2)If the magnetic field strength in the spectrometer is increased, will the spacing between where the 235mU and 238mU ions strike increase, decrease, or stay the same?

3)If the initial velocity of the ions as they enter the spectrometer is increased, will the spacing between where the 235mU and 238mU ions strike increase, decrease, or stay the same?

4)If a beam of singly ionized carbon atoms (12mC and 14mC) is injected into the same mass spectrometer as the uranium ions, will the spacing between the carbon ions be greater than, less than, or equal to the spacing between the uranium ions?

Answer 1

Concepts and reason

The concept required to solve the given problem is magnetic force and centripetal force.

Equate the centripetal and magnetic force to determine the equation for the spacing between where the $^{235}U{\rm{ and}}{{\rm{ }}^{238}}U$ will strike and then determine the dependence of initial velocity and magnetic field on it.

Fundamentals

The magnetic force is given by,

$\begin{array}{c}\\{{\vec F}_M} = q\left( {\vec v \times \vec B} \right)\\\\ = qvB\sin \theta {\rm{ \hat n}}\\\end{array}$

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

The centripetal force is the force that acts on a body when it moves along a circle and it is always directed towards the center.

(1)

The magnetic force is given by,

$\begin{array}{c}\\{{\vec F}_M} = q\left( {\vec v \times \vec B} \right)\\\\ = qvB\sin \theta {\rm{ \hat n}}\\\end{array}$

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

The centripetal force is given by,

${F_C} = \frac{{m{v^2}}}{R}$

Here, $m$ is the mass, $v$ is the velocity and $R$ is the radius of the path traced.

Since the $^{238}U$ proton moves in a circular path, the necessary centripetal force will be provided by the magnetic force.

Thus,

${F_C} = {F_M}$

Substitute $\frac{{m{v^2}}}{R}$ for ${F_C}$ and $qvB$ for ${F_M}$ in the above equation.

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

The radius of the path traced by $^{238}U$ will be,

${R_{^{238}U}} = \frac{{{m_{^{238}U}}v}}{{qB}}$

The radius of the path traced by $^{235}U$ will be,

${R_{^{235}U}} = \frac{{{m_{^{235}U}}v}}{{qB}}$

Since ${m_{^{238}U}} > {m_{^{235}U}}$ , the radius of $^{238}U$ will be greater than $^{235}U$ .

Thus, ${R_{^{238}U}} > {R_{^{235}U}}$ and hence $^{235}U$ will strike the collecting plate above $^{238}U$ .

(2)

The spacing between where the $^{235}U{\rm{ and}}{{\rm{ }}^{238}}U$ will strike will be given by,

${R_{^{238}U}} - {R_{^{235}U}} = \frac{{{m_{^{238}U}}v}}{{qB}} - \frac{{{m_{^{235}U}}v}}{{qB}}$

From the above equation, the spacing is inversely proportional to the magnetic field.

Hence if the magnetic field is increased the spacing will decrease.

(3)

The spacing between where the $^{235}U{\rm{ and}}{{\rm{ }}^{238}}U$ will strike will be given by,

${R_{^{238}U}} - {R_{^{235}U}} = \frac{{{m_{^{238}U}}v}}{{qB}} - \frac{{{m_{^{235}U}}v}}{{qB}}$

Rearrange the above equation.

$\left( {{R_{^{238}U}} - {R_{^{235}U}}} \right) = \frac{{\left( {{m_{^{238}U}} - {m_{^{235}U}}} \right)v}}{{qB}}$

From the above equation, the spacing is directly proportional to the initial velocity of the ions.

Hence if the initial velocity of the ions is increased the spacing will also increase.

(4)

The spacing between where the $^{14}{\rm{C and}}{{\rm{ }}^{12}}{\rm{C}}$ will strike will be given by,

${R_{^{14}C}} - {R_{^{12}C}} = \frac{{{m_{^{14}C}}v}}{{qB}} - \frac{{{m_{^{12}C}}v}}{{qB}}$

Rearrange the above equation.

$\left( {{R_{^{14}C}} - {R_{^{12}C}}} \right) = \frac{{\left( {{m_{^{14}C}} - {m_{^{12}C}}} \right)v}}{{qB}}$

Since the mass difference between $^{14}{\rm{C and}}{{\rm{ }}^{12}}{\rm{C}}$ is less than the mass difference between $^{235}U{\rm{ and}}{{\rm{ }}^{238}}U$ therefore the spacing between the carbon ions will be less than that for the uranium ions.

Ans: Part 1

The $^{235}U$ will strike the collecting plate above $^{238}U$ .