91Ó°ÊÓ

The force between the proton and uranium nucleus is due to the Coulombic repulsion between these two positive charges.

Mass of the proton is,$1.67×{10}^{-27}\mathrm{kg}$

The initial speed of the proton is, ${\mathrm{v}}_0=3.00×{10}^5\mathrm{m}/\mathrm{s}$

Uranium nucleus is at 5.0 m from the proton

Repelling force of proton is, $\mathrm{F}=\frac{\mathrm{Î\pm }}{{\mathrm{x}}^2}$

Here, x is the separation between the two objects and $\mathrm{Î\pm }=2.12×{10}^{26}\mathrm{N}·{\mathrm{m}}^2$

Work done by the force when the proton is$8.00×{10}^{-10}$from the Uranium nucleus,

$$\begin{gathered} \mathrm{W}={∫}_{{\mathrm{x}}_0}^{\mathrm{x}}\left(\frac{\mathrm{Î\pm }}{{\mathrm{x}}^2}\right)\mathrm{dx} \\ ={\left[\frac{-\mathrm{Î\pm }}{\mathrm{x}}\right]}_{{\mathrm{x}}_0}^{\mathrm{x}} \\ ={\left[\frac{-2.12×{10}^{-26}}{\mathrm{x}}\right]}_{5.0\mathrm{m}}^{8.0×{10}^{-10}\mathrm{m}} \\ =-2.56×{10}^{-17}\mathrm{J} \end{gathered}$$

Now, using the Work-Energy theorem, we get

$$\begin{gathered} \mathrm{W}={\mathrm{K}}_2-{\mathrm{K}}_1 \\ -2.65×{10}^{-17}\mathrm{J}=\frac{1}{2}{\mathrm{mv}}^2-\frac{1}{2}{\mathrm{mv}}_0^2 \\ -2.65×{10}^{-17}\mathrm{J}=\frac{1}{2}\left(1.67×{10}^{-27}\mathrm{kg}\right)\left[{\mathrm{v}}^2-\left(3.00×{10}^5\mathrm{m}·{\mathrm{s}}^{-2}\right)\right] \\ \mathrm{v}=2.414×{10}^5\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the speed of the proton is $2.414×{10}^5\mathrm{m}/\mathrm{s}$.

Applying the Work-Energy theorem again (but with$\mathrm{v}=0\mathrm{m}/\mathrm{s}$this time because the proton is momentarily at its closest distance to the uranium nucleus), we can find how close the uranium nucleus gets.

$$\begin{gathered} \mathrm{W}={\mathrm{K}}_2-{\mathrm{K}}_1 \\ ∫\mathrm{Fdx}={\mathrm{K}}_2-{\mathrm{K}}_1 \\ ∫\left(\frac{\mathrm{Î\pm }}{{\mathrm{x}}^2}\right)\mathrm{dx}=\frac{1}{2}{\mathrm{mv}}^2-\frac{1}{2}{\mathrm{mv}}_0^2 \\ -\frac{\mathrm{Î\pm }}{\mathrm{x}}=-\frac{1}{2}{\mathrm{mv}}_0^2 \\ {\mathrm{x}}_{\min }=\frac{2\left(2.12×{10}^{-26}\mathrm{N}·{\mathrm{m}}^2\right)}{\left(1.67×{10}^{-27}\mathrm{kg}\right)\left(3.00×{10}^5\mathrm{m}/\mathrm{s}\right)} \\ {\mathrm{x}}_{\min }=2.82×{10}^{-10}\mathrm{m} \end{gathered}$$

Hence, the closest distance to the uranium nucleus is $2.82×{10}^{-10}\mathrm{m}$.

Using the Work-Energy theorem yet again with the initial velocity equal to zero (when the proton is momentarily at rest), we can find the final velocity of the proton when it is again 5.0 m away from the nucleus.

$$\begin{gathered} \mathrm{W}={\mathrm{K}}_2-{\mathrm{K}}_1 \\ ∫\mathrm{Fdx}={\mathrm{K}}_2-{\mathrm{K}}_1 \\ {∫}_{2.82×{10}^{-10}\mathrm{m}}^{5.0\mathrm{m}}\left(\frac{\mathrm{Î\pm }}{{\mathrm{x}}^2}\right)\mathrm{dx}=\frac{1}{2}{\mathrm{mv}}^2-\frac{1}{2}{\mathrm{mv}}_0^2 \\ {\left[-\frac{\mathrm{Î\pm }}{\mathrm{x}}\right]}_{2.82×{10}^{-10}\mathrm{m}}^{5.0\mathrm{m}}=\frac{1}{2}{\mathrm{mv}}_{\mathrm{f}}^2 \\ \left[\frac{\left(-2.12×{10}^{-26}\mathrm{N}·{\mathrm{m}}^2\right)}{5.0\mathrm{m}}-\frac{\left(-2.12×{10}^{-26}\mathrm{N}·{\mathrm{m}}^2\right)}{2.82×{10}^{-10}\mathrm{m}}\right]=\left(0.5\right)×\left(1.67×{10}^{-27}\mathrm{kg}\right){\mathrm{v}}_{\mathrm{f}}^2 \\ {\mathrm{v}}_{\mathrm{f}}=3.00×{10}^5\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the speed of the proton is $3.00×{10}^5\mathrm{m}/\mathrm{s}$when it is again 5.0 m .