Chapter 19, Problem 128P

(a)

To determine

The speed with which the ${}^{238}{\text{U}}^{2+}$ ions emerge from the accelerating plates.

Answer to Problem 128P

The speed with which the ${}^{238}{\text{U}}^{2+}$ ions emerge from the accelerating plates is $\sqrt{2}v$ .

Explanation of Solution

The kinetic energy gained by the ion must be equal to the potential energy lost.

Write the equation for the conservation of energy.

  $\Delta K=\Delta U$        (I)

Here, $\Delta K$ is the gain kinetic energy of the ion and $\Delta U$ is the loss in potential energy by the ion.

Write the equation for $\Delta K$ .

  $\Delta K=\frac{1}{2}m{v}^2$        (II)

Here, $m$ is the mass of the ion.

Write the equation for $\Delta U$ .

  $\Delta U=q\Delta V$        (III)

Here, $q$ is the charge of the ion and $\Delta V$ is the potential difference through which the ion is accelerated.

Put equations (II) and (III) in equation (I).

  $\begin{array}{c}\frac{1}{2}m{v}^2=q\Delta V\\ v=\sqrt{\frac{2q\Delta V}{m}}\\ v\propto \sqrt{q}\\ v=k\sqrt{q}\end{array}$        (IV)

Here, $k$ is the proportionality constant.

Write the relation for the speed of ${}^{235}{\text{U}}^{+}$ ions using relation (IV).

  $v_{2+}=k\sqrt{q_{2+}}$

Here, $v_{2+}$ is the speed of the ${}^{238}{\text{U}}^{2+}$ ions and $q_{2+}$ is the charge of the ${}^{238}{\text{U}}^{2+}$ ions.

Write the relation for the speed of ${}^{238}{\text{U}}^{+}$ ions using relation (IV).

  $v_{+}=k\sqrt{q_{+}}$

Here, $v_{+}$ is the speed of the ${}^{238}{\text{U}}^{+}$ ions and $q_{+}$ is the charge of the ${}^{238}{\text{U}}^{+}$ ions.

Take the ratio of the above two equations.

  $\begin{array}{c}\frac{v_{2+}}{v_{+}}=\frac{k\sqrt{q_{2+}}}{k\sqrt{q_{+}}}\\ =\sqrt{\frac{q_{2+}}{q_{+}}}\\ v_{2+}=\sqrt{\frac{q_{2+}}{q_{+}}}{v}_{+}\end{array}$        (V)

Conclusion:

The charge of the ion ${}^{238}{\text{U}}^{+}$ ions is equal to $+e$ and that of ${}^{238}{\text{U}}^{2+}$ ions is $+2e$ where $e$ is the magnitude of electronic charge.

Substitute $+2e$ for $q_{2+}$ , $+e$ for $q_{+}$ and $v$ for $v_{+}$ in equation (V) to find $v_{2+}$ .

  $\begin{array}{c}{v}_{2+}=\sqrt{\frac{+2e}{+e}}v\\ =\sqrt{2}v\end{array}$

Therefore, the speed with which the ${}^{238}{\text{U}}^{2+}$ ions emerge from the accelerating plates is $\sqrt{2}v$ .

(b)

To determine

The sketch of the trajectory of ${}^{238}{\text{U}}^{2+}$ ions inside the velocity selector.

Answer to Problem 128P

The sketch of the trajectory of ${}^{238}{\text{U}}^{2+}$ ions inside the velocity selector is

Explanation of Solution

In a velocity selector, for an ion with speed $v$ , the electric and magnetic fields will be equal in magnitude and opposite in direction. For an ion with speed greater than $v$ , the magnetic force will dominate. In part (a), it is found that the speed of the ${}^{238}{\text{U}}^{2+}$ ions is $\sqrt{2}v$ which is greater than $v$ . This implies for ${}^{238}{\text{U}}^{2+}$ ions magnetic force dominates.

Write the equation for the magnetic force.

  ${\overrightarrow{F}}_B=q\overrightarrow{v}\times \overrightarrow{B}$        (VI)

Here, ${\overrightarrow{F}}_B$ is the magnetic force, $\overrightarrow{v}$ is the velocity of the charge and $\overrightarrow{B}$ is the magnetic field.

In the velocity selector, ${\overrightarrow{F}}_E$ must oppose ${\overrightarrow{F}}_B$ . The direction of the magnetic force will be perpendicular to both velocity vector and magnetic field according to equation. In accordance with the right-hand rule, ${\overrightarrow{F}}_B$ will be directed downward for a positive ion. The sketch of the ${}^{238}{\text{U}}^{2+}$ ions inside the velocity selector is shown below.

Conclusion:

Thus, the sketch of the ${}^{238}{\text{U}}^{2+}$ ions inside the velocity selector is sketched in figure 1.

(c)

To determine

The diameter of the path of ${}^{238}{\text{U}}^{2+}$ ions.

Answer to Problem 128P

The diameter of the path of ${}^{238}{\text{U}}^{2+}$ ions is $\frac{D}{\sqrt{2}}$ .

Explanation of Solution

Write the equation for the magnitude of the magnetic force on ions using equation (V).

  $F_B=qvB$        (VI)

Here, $F_B$ is the magnitude of the magnetic force and $B$ is the magnitude of the magnetic field.

The magnetic force provides the centripetal force needed for the circular motion of the ions.

Write the relationship between the magnetic force and the centripetal force.

  $F_B=F_C$        (VII)

Here, $F_C$ is the magnitude of the centripetal force.

Write the equation for the magnitude of the centripetal force on ions.

  $F_C=\frac{m{v}^2}{r}$        (VIII)

Here, $r$ is the radius of the circular path.

Write the equation for the radius of a circle.

  $r=\frac{D}{2}$

Here, $D$ is the diameter of the circular path.

Put the above equation in equation (VIII).

  $\begin{array}{c}{F}_C=\frac{m{v}^2}{\left(\frac{D}{2}\right)}\\ =m\frac{2v^2}{D}\end{array}$        (IX)

Put equations (VI) and (IX) in equation (VII).

  $\begin{array}{c}qvB=m\frac{2v^2}{D}\\ \frac{B}{2m}=\frac{v}{qD}\\ \text{constant}=\frac{v}{qD}\end{array}$

Since $m$ and $B$ are constants for the two kinds of ions, the left-hand side of the above equation must be constant. This implies the right-hand side of the above equation is also constant.

Write the relationship between the diameters of the paths of ${}^{238}{\text{U}}^{+}$ ions and ${}^{238}{\text{U}}^{2+}$ ions using the above relation.

  $\frac{v}{qD}=\frac{v_{2+}}{q_{2+}{D}_{2+}}$

Here, $v$ is the speed of the ${}^{238}{\text{U}}^{+}$ ions, $D$ is the diameter of the circular path of ${}^{238}{\text{U}}^{+}$ ions , $q$ is the charge of the ${}^{238}{\text{U}}^{+}$ ions and $D_{2+}$ is the diameter of the circular path of ${}^{238}{\text{U}}^{2+}$ ions.

Rewrite the above relationship for $D_{2+}$ .

  $D_{2+}=\frac{v_{2+}q}{v{q}_{2+}}D$        (X)

Conclusion:

Substitute $\sqrt{2}v$ for $v_{2+}$ , $+e$ for $q$ and $+2e$ for $q_{2+}$ in equation (X) to find $D_{2+}$ .

  $\begin{array}{c}{D}_{2+}=\frac{\left(\sqrt{2}v\right)\left(+e\right)}{v\left(+2e\right)}D\\ =\frac{\sqrt{2}}{2}D\\ =\frac{D}{\sqrt{2}}\end{array}$

Therefore, the diameter of the path of ${}^{238}{\text{U}}^{2+}$ ions is $\frac{D}{\sqrt{2}}$ .