Answer:
[tex]3.95 \times 10^{-14}m[/tex]
Explanation:
We are given that
Charged on alpha particle=q=2e=[tex]2\times 1.6\times 10^{-19} C[/tex]
Where [tex]e=1.6\times 10^{-19} C[/tex]
Initial kinetic energy=K.E=5.76 MeV=[tex]5.76\times 10^6\times 1.6\times 10^{-9} C[/tex]
[tex]1 Me V=10^6\times 1.6\times 10^{-19} V[/tex]
Z=79
Charge on protons=[tex]q'=79\times 1.6\times 10^{-19} C[/tex]
We have to find the closeness of alpha particle to the gold nucleus before being turned around.
Initial kinetic energy=Final potential energy
[tex]5.76\times 10^6\times 1.6\times 10^{-9}=\frac{Kq_1q_2}{r}[/tex]
Where [tex]k=9\times 10^9[/tex]
[tex]r=\frac{9\times 10^9\times 79\times 1.6\times 10^{-19}\times 2\times 1.6\times 10^{-19}}{5.76\times 10^6\times 1.6\times 10^{-9}}[/tex]
[tex]r=3.95\times 10^{-14}m[/tex]
