Chapter 17, Problem 119P

(a)

To determine

The minimum distance between the proton and the nucleus.

Answer to Problem 119P

The minimum distance between the proton and the nucleus is $\underset{\_}{3.48\times {10}^{-12}\text{m}}$.

Explanation of Solution

Write the expression for the law of conservation of momentum.

    $m_p{v}_i=m_{\text{Li}}{v}_{\text{Li}}$

Here, $m_p$ is the mass of the proton, $v_i$ is the initial velocity of the proton, $m_{\text{Li}}$ is the mass of Lithium nucleus and $v_{\text{Li}}$ is the velocity of the Lithium nucleus.

Rewrite the above equation to find the velocity of the Lithium nucleus.

    $v_{\text{Li}}=\frac{m_p{v}_i}{m_{\text{Li}}}$                                                                                                                (I)

Write the expression for the law of conservation of energy.

    $\frac{1}{2}{m}_p{v}_i^2=\frac{1}{2}{m}_{\text{Li}}{v}_{\text{Li}}^2+\frac{k{q}_1{q}_2}{r}$

Here, $k$ is a constant, $q_1$ is the charge of the proton, $q_2$ is the charge of the nucleus and $r$ is the separation distance.

Rewrite the above equation to find the minimum distance between the proton and the nucleus.

    $r=\frac{k{q}_1{q}_2}{\left(\frac{1}{2}{m}_p{v}_i^2-\frac{1}{2}{m}_{\text{Li}}{v}_{\text{Li}}^2\right)}$                                                                                          (II)

Conclusion:

Substitute $1.007\text{u}$ for $m_p$, $5.24\times {10}^5\text{m}/\text{s}$ for $v_i$, and $7.014\text{u}$ for $m_{\text{Li}}$ in equation (I).

    $\begin{array}{c}{v}_{\text{Li}}=\frac{\left(1.007\text{u}\right)\left(5.24\times {10}^5\text{m}/\text{s}\right)}{7.014\text{u}}\\ =7.523\times {10}^4\text{m}/\text{s}\end{array}$

Substitute $8.988\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2$ for $k$, $3e$ for $q_1{q}_2$, $1.007\text{u}$ for $m_p$, $5.24\times {10}^5\text{m}/\text{s}$ for $v_i$, $7.014\text{u}$ for $m_{\text{Li}}$ and $7.523\times {10}^4\text{m}/\text{s}$ for $v_{\text{Li}}$ in equation (II).

    $\begin{array}{c}r=\frac{\left(8.988\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)\left(3e\right)}{\left(\frac{1}{2}\left(1.007\text{u}\right){\left(5.24\times {10}^5\text{m}/\text{s}\right)}^2-\frac{1}{2}\left(7.014\text{u}\right){\left(7.523\times {10}^4\text{m}/\text{s}\right)}^2\right)}\\ =3.48\times {10}^{-12}\text{m}\end{array}$

Therefore, the minimum distance between the proton and the nucleus is $\underset{\_}{3.48\times {10}^{-12}\text{m}}$.

(b)

To determine

The recoil speed of the nucleus.

Answer to Problem 119P

The recoil speed of the nucleus is $\underset{\_}{132\text{km}/\text{s}}$.

Explanation of Solution

Write the expression for momentum conservation when particles are separated by large distances.

    $m_p{v}_i=m_p{v}_{1f}+m_{\text{Li}}{v}_{2f}$                                                                                            (III)

Write the expression for velocities after collision.

    $-v_i=-\left(v_{2f}-v_{1f}\right)$                                                                                                  (IV)

Conclusion:

Substitute $524\text{km}/\text{s}$ for $v_i$, $1.007\text{u}$ for $m_p$ and $7.014\text{u}$ for $m_{\text{Li}}$ in equation (III).

    $\left(1.007\text{u}\right)\left(524\text{km}/\text{s}\right)=\left(1.007\text{u}\right)v_{1f}+\left(7.014\text{u}\right)v_{2f}$                                               (V)

Substitute $524\text{km}/\text{s}$ for $v_i$ in equation (IV).

    $-\left(524\text{km}/\text{s}\right)=-\left(v_{2f}-v_{1f}\right)$                                                                                   (VI)

Solve equations (V) and (VI) to find the speed of the recoil nucleus.

    $v_{2f}=132\text{km}/\text{s}$

Therefore, the recoil speed of the nucleus is $\underset{\_}{132\text{km}/\text{s}}$.