91Ó°ÊÓ

The potential energy U between two charges $q_1$and $q_2$at a distance of separation of r is expressed as, $U=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{q_1{q}_2}{r}$.

Here the case is a nucleus and alpha particle. The nucleus can be considered as a sphere with all of its positive charge $Q=+79e$concentrated at its center. Then $q_1=Q$, and charge of alpha particle be$q_2=q$.The distance is considered when alpha particle is $d=2.0*{10}^{-14}\mathrm{m}$close to the surface. Then, the total distance of separation localid="1664281613185" $r$is the sum of the radius of gold nucleus = $7.3*{10}^{-15}\mathrm{m}$and the distance outside the nucleus $d$. Then, $\mathrm{r}=7.3*{10}^{-15}+20*{10}^{(}-15)=27.3*{10}^{-15}\mathrm{m}$.

The potential energy expression can be rearranged$=27.3*{10}^{-15}\mathrm{m}$.

Since the potential energy is equivalent to the work done, the potential can be defined as the work done per unit charge. $V=\frac{W}{q}=\frac{U}{q}$. Thus, potential energy expression is $U=qV$.

Comparing the two expressions,

$qV=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{Qq}{R+d}$, The potential equation is $V=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{Q}{r}$.

Substituting the values of ${\mathrm{Î\mu }}_0=8.854*10-12m-3kg-1s^4{A}^2$, $q=+2e=2*1.6*{10}^{-19}C$$Q=+79e=+79*1.6*{10}^{-19}C$, $r=27.3*{10}^{-15}\mathrm{m}$in localid="1664282221657" $V=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{Q}{r}$,

$$\begin{gathered} V=(9*{10}^9)\frac{79*1.6*{10}^{-19}}{27.3*{10}^{-15}} \\ V=4.167*{10}^6=4.2*{10}^{6}V \end{gathered}$$

Hence, the voltage up to which the alpha particle must be accelerated is $4.2*{10}^{6}V$.