$FIGUREQ25.2$shows the potential energy of a proton $\left(q=+e\right)$and a lead nucleus$\left(q=+82e\right)$. The horizontal scale is in units of femtometers, where $1fm={10}^{-15}m.$

$\left(a\right)$A proton is fired toward a lead nucleus from very far away. How much initial kinetic energy does the proton need to reach a turning point$10fm$ from the nucleus? Explain.
$\left(b\right)$How much kinetic energy does the proton of part a have when it is$20fm$from the nucleus and moving toward it, before the collision?

The electron's potential energy is more than its kinetic energy, yet it is negative. With increasing orbit radii, the electron's potential energy drops. The kinetic energy of a proton and an electron are the same. Because the mass of the proton is more than that of the electron, we may argue that the proton has more energy.

(A) There an initial kinetic energy need for a proton to reach a turning point from the nucleus

The electric force is cautious,so

$$\begin{gathered} \mathrm{\Delta }K=-\mathrm{\Delta }U, \\ {K}_f-K_i=U_i-U_f. \end{gathered}$$

Having that,

$K_f=0$and $U_i=0.$

Thus

localid="1648740171630" $K_i=U_f$

At localid="1648739440815" $10fm$,

localid="1648740178686" $U_f=2\times {10}^{-12}J.$

(B) Kinetic energy does a proton in part a have when it is travelling away from the nucleus and toward it before colliding

At $20fm$,

localid="1648739541885" $\mathrm{K}+\mathrm{U}=$total energylocalid="1648739575918" $=2\times {10}^{-12}\mathrm{J}.$

From the graph,

localid="1650237867915" $$\begin{gathered} U\left(20\mathrm{fm}\right)=1\times {10}^{-12}\mathrm{J} \\ \mathrm{K}+1\times {10}^{-12}\mathrm{J} \\ =2\times {10}^{-12}\mathrm{J}. \end{gathered}$$