Chapter 29, Problem 63AP

To determine

The radius of the alpha particle’s trajectory.

Answer to Problem 63AP

The radius of the alpha particle’s trajectory is $\frac{3R}{4}$.

Explanation of Solution

Write the equation for the radius of the trajectory of the proton.

    $R=\frac{m_{\text{p}}{v}_1}{B{q}_1}$                                                                                                                   (I)

Here, $R$ is the radius of the trajectory of the proton, $m_{\text{p}}$ is the mass of the proton, $v_1$ is the velocity of the proton after the collision, $B$ is the magnetic field and $q_1$ is the charge on the proton.

Write the equation for the radius of the trajectory of the alpha particle.

    $R^{\prime }=\frac{m_{\text{a}}{v}_2}{B{q}_2}$                                                                                                                 (II)

Here, $R^{\prime }$ is the radius of the trajectory of the alpha particle, $m_{\text{a}}$ is the mass of the alpha particle, $v_2$ is the velocity of the alpha particle after the collision and $q_2$ is the charge on the alpha particle.

Write the equation for the principle of conservation of linear momentum.

    $m_{\text{a}}v=m_{\text{p}}{v}_1+m_{\text{a}}{v}_2$                                                                                                    (III)

The alpha particle's mass is four times the proton's mass.

    $m_{\text{a}}=4m_{\text{p}}$

Substitute $4m_{\text{p}}$ for $m_{\text{a}}$ in the equation (III).

    $\begin{array}{c}4m_{\text{p}}v=m_{\text{p}}{v}_1+4m_{\text{p}}{v}_2\\ 4v=v_1+4v_2\end{array}$                                                                                                (IV)

Write the equation for the principle of conservation of energy.

    $\frac{1}{2}{m}_{\text{a}}{v}^2=\frac{1}{2}{m}_{\text{p}}{v}_1^2+\frac{1}{2}{m}_{\text{a}}{v}_2^2$                                                                                        (V)

Substitute $4m_{\text{p}}$ for $m_{\text{a}}$ in the above equation.

    $\begin{array}{c}\frac{1}{2}\left(4m_{\text{p}}\right)v^2=\frac{1}{2}{m}_{\text{p}}{v}_1^2+\frac{1}{2}\left(4m_{\text{p}}\right)v_2^2\\ 4v^2=v_1^2+4v_2^2\end{array}$                                                                            (VI)

Take the square of equation (IV) and then divide by $4$.

    $\begin{array}{c}\frac{{\left(4v\right)}^2}{4}=\frac{{\left(v_1+4v_2\right)}^2}{4}\\ 4v^2=\frac{{\left(v_1+4v_2\right)}^2}{4}\end{array}$                                                                                                (VII)

Compare the equation (VI) and (VII).

    $\begin{array}{c}{v}_1^2+4v_2^2=\frac{{\left(v_1+4v_2\right)}^2}{4}\\ 4v_1^2+16v_2^2=v_1^2+16v_2^2+8v_1{v}_2\\ 3v_1^2=8v_1{v}_2\\ 3v_1=8v_2\end{array}$                                                                               (VIII)

Rewrite the equation (I).

    $v_1=\frac{B{q}_{1}R}{m_{\text{p}}}$

Rewrite the equation (II).

    $v_2=\frac{B{q}_2{R}^{\prime }}{m_{\text{a}}}$

Substitute $\frac{B{q}_{1}R}{m_{\text{p}}}$ for $v_1$ and $\frac{B{q}_2{R}^{\prime }}{m_{\text{a}}}$ for $v_2$ in the equation (VIII).

    $3\left(\frac{B{q}_{1}R}{m_{\text{p}}}\right)=8\left(\frac{B{q}_2{R}^{\prime }}{m_{\text{a}}}\right)$

The alpha particle's charge is twice the proton's charge.

    $q_2=2q_1$

Substitute $2q_1$ for $q_2$ and $4m_{\text{p}}$ for $m_{\text{a}}$ in the above equation.

    $\begin{array}{c}3\left(\frac{B{q}_{1}R}{m_{\text{p}}}\right)=8\left(\frac{B\left(2q_1\right)R^{\prime }}{4m_{\text{p}}}\right)\\ 3R=4R^{\prime }\\ R^{\prime }=\frac{3R}{4}\end{array}$

Conclusion:

Therefore, the radius of the alpha particle’s trajectory is $\frac{3R}{4}$.