Question

In experiments in which atomic nuclei collide, head-on collisions like that described in Problem 23.74 do happen, but “near misses” are more common. Suppose the alpha particle in that problem is not “aimed” at the center of the lead nucleus but has an initial nonzero angular momentum (with respect to the stationary lead nucleus) of magnitude L = p₀b, where p₀ is the magnitude of the particle’s initial momentum and b = 1.00 x 10-12 m. What is the distance of closest approach? Repeat for b = 1.00 x 10-13 m and b = 1.00 x 10-14 m.

Short answer

The distance of closest approach when b = 1.00 x 10-12m is r1=1.01 x 10-12m.

The distance of closest approach when b = 1.00 x 10-13m is r2=1.11 x 10-13m

The distance of closest approach when b = 1.00 x 10-14 m is r3=2.54 x 10-14 m

Step-by-step solution

Step 1:

According to law of conservation of energy

$$\begin{gathered} \mathbf{\Delta KE}\mathbf{-}\mathbf{\Delta U}\mathbf{=}\mathbf{0} \\ {\mathbf{K}}_{\mathbf{a}}\mathbf{+}{\mathbf{U}}_{\mathbf{a}}\mathbf{=}{\mathbf{K}}_{\mathbf{b}}\mathbf{+}{\mathbf{U}}_{\mathbf{b}} \end{gathered}$$

Where Ua=0

${\mathrm{K}}_{\mathrm{a}}=\frac{1}{{\mathrm{r}}_{\mathrm{i}}}{{\mathrm{mv}}_{\mathrm{b}}}^2+\frac{{\mathrm{kq}}_1{\mathrm{q}}_2}{{\mathrm{r}}_{\mathrm{i}}}$

According to law of conservation of momentum

localid="1664883029399" ${\mathbf{P}}_{\mathbf{a}}\mathbf{=}{\mathbf{P}}_{\mathbf{b}}$

$$\begin{gathered} {\mathrm{mv}}_{\mathrm{a}}\mathrm{b}={\mathrm{mv}}_{\mathrm{b}}{\mathrm{r}}_{\mathrm{i}} \\ \frac{{\mathrm{v}}_{\mathrm{a}}{\mathrm{b}}_{\mathrm{i}}}{{\mathrm{r}}_{\mathrm{i}}}={\mathrm{v}}_{\mathrm{b}} \end{gathered}$$

Substitute to get Ka

$$\begin{gathered} {\mathrm{K}}_{\mathrm{a}}=\frac{1}{2}{{\mathrm{mv}}^2}_{\mathrm{b}}+\frac{{\mathrm{kq}}_1{\mathrm{q}}_2}{{\mathrm{r}}_{\mathrm{i}}} \\ =\frac{1}{2}\mathrm{m}{\left(\frac{{\mathrm{v}}_{\mathrm{a}}\mathrm{b}}{{\mathrm{r}}_{\mathrm{i}}}\right)}^2+\frac{{\mathrm{kq}}_1{\mathrm{q}}_2}{{\mathrm{r}}_{\mathrm{i}}} \\ =\frac{1}{2}{{\mathrm{mv}}^2}_{\mathrm{b}}{\left(\frac{\mathrm{b}}{{\mathrm{r}}_{\mathrm{i}}}\right)}^2+\frac{{\mathrm{kq}}_1{\mathrm{q}}_2}{{\mathrm{r}}_{\mathrm{i}}} \\ ={\mathrm{K}}_{\mathrm{a}}{\left(\frac{\mathrm{b}}{{\mathrm{r}}_{\mathrm{i}}}\right)}^2+\frac{{\mathrm{kq}}_1{\mathrm{q}}_2}{{\mathrm{r}}_{\mathrm{i}}} \\ {\mathrm{K}}_{\mathrm{a}}{{\mathrm{r}}_{\mathrm{i}}}^2-{\mathrm{kq}}_1{\mathrm{q}}_2{\mathrm{r}}_{\mathrm{i}}-{\mathrm{K}}_{\mathrm{a}}{\mathrm{b}}^2\mathrm{i}=0 \end{gathered}$$

Step 2:

Substitute to get value of r1at b1:

$[1.76\times {10}^{-12}]{{\mathrm{r}}_1}^2-3.78\times {10}^{-26}{\mathrm{r}}_1-1.76\times {10}^{-12}{(1.00\times {10}^{-12})}^2=0$

Solve the above equation to get r1:

$$\begin{gathered} {\mathrm{r}}_1=\frac{-\mathrm{b}\pm \sqrt{{\mathrm{b}}^2-4\mathrm{ac}}}{2\mathrm{a}} \\ =\frac{-(-3.78\times {10}^{-26})\pm \sqrt{\left(-3.78\times {10}^{-26}\right)-4\left[1.76\times {10}^{-12}\right]}\left[1.76\times {10}^{-12}\left(1.00\times {10}^{-12}\right)\right]}{2\left[1.76\times {10}^{-12}\right]} \\ =1.01\times {10}^{-12}\mathrm{m} \end{gathered}$$

Therefore, the distance of closest approach when b = 1.00 x 10-12 m is r1=1.01 x 10-12 m.

Step 3:

Substitute to get value of r2at b2:

$[1.76\times {10}^{-12}]{{\mathrm{r}}_1}^2-3.78\times {10}^{-26}{\mathrm{r}}_1-1.76\times {10}^{-12}{(1.00\times {10}^{-13})}^2=0$

Solve the above equation to get r2:

$$\begin{gathered} \mathrm{r}2=\frac{-\mathrm{b}\pm \sqrt{{\mathrm{b}}^2-4\mathrm{ac}}}{2\mathrm{a}} \\ =\frac{-(-3.78\times {10}^{-26})\pm \sqrt{\left(-3.78\times {10}^{-26}\right)-4\left[1.76\times {10}^{-12}\right]}\left[1.76\times {10}^{-12}\left(1.00\times {10}^{-12}\right)\right]}{2\left[1.76\times {10}^{-12}\right]} \\ =1.11\times {10}^{-13}\mathrm{m} \end{gathered}$$

Therefore, the distance of closest approach when b = 1.00 x 10-13 m is r2=1.11 x 10-13 m.

Step 4:

Substitute to get value of r3at b3:

$[1.76\times {10}^{-12}]{{\mathrm{r}}_1}^2-3.78\times {10}^{-26}{\mathrm{r}}_1-1.76\times {10}^{-12}{(1.00\times {10}^{-13})}^2=0$

Solve the above equation to get r3:

$$\begin{gathered} {\mathrm{r}}_3=\frac{-\mathrm{b}\pm \sqrt{{\mathrm{b}}^2-4\mathrm{ac}}}{2\mathrm{a}} \\ =\frac{-(-3.78\times {10}^{-26})\pm \sqrt{\left(-3.78\times {10}^{-26}\right)-4\left[1.76\times {10}^{-12}\right]}\left[1.76\times {10}^{-12}\left(1.00\times {10}^{-12}\right)\right]}{2\left[1.76\times {10}^{-12}\right]} \\ =2.54\times {10}^{-14}\mathrm{m} \end{gathered}$$

Therefore, the distance of closest approach when b = 1.00 x 10-14 m is r3=2.54 x 10-14 m.