Answer:
Explanation:
a) The total energy at closest approach is the kinetic energy + potential energy at that point. Let [tex]v_0[/tex] be the velocity and [tex]r_0[/tex] the distance at the closest approach.
The total energy there is then
[tex]PE =\frac{K\times q_1\times q_2}{r_0}
\\\\KE = 0.5\times m\times v_0^2[/tex]
Total energy
[tex]U(r_0) =\frac{K\times q_1\times q_2}{r_0} + 0.5\times m\times v_0^2[/tex]
Using conservation of angular momentum,
[tex]initial = v_{\infty}\times m\times b
[/tex]
at [tex]r_0 = v_0\times m\times r_0[/tex]
These must be equal so
[tex]v_{\infty}\times m\times b = v_0\times m\times r_0\\v_0=\frac{v_{\infty}\times b}{r_0}
[/tex]
substitute this for tex]v_0[/tex] in the energy equation
[tex]U(r_0) = \frac{K\times q_1\times q_2}{r_0} + \frac{0.5\times m\times v_{\infty}^2\times b^2}{r_0^2}
[/tex]
This must equal the initial kinetic energy [tex]U_{\infty}= (11 MeV)[/tex]
[tex]U_\infty= \frac{K\times q_1\times q_2}{r_0} + \frac{0.5\times m\times v_{\infty}^2\times b^2}{r_0^2}[/tex]
however, [tex]0.5\times m\times v_{\infty}^2 = U_{\infty}[/tex]
[tex]U_\infty= \frac{K\times q_1\times q_2}{r_0} + \frac{U_{\infty}\times b^2}{r_0^2}[/tex]
[tex]U_{\infty}\times r_0^2 - (K\times q_1\times q_2)\times r_0 - U_{\infty}\times b^2 = 0\\\\r_0^2 - (\frac{K\times q_1\times q_2}{U_{\infty})\times r_0 - b^2 = 0[/tex]
[tex]U_{\infty} = 11\times 10^6\times 1.6\times 10^{-19} J = 1.76\times 10^{-12} J[/tex]
[tex]q_1 = 2e = 3.2\times 10^{-19} C\\q_2 = 82e = 131.2\times 10^{-19} C\\K\times q_1\times q_2 = 3.78\times 10^{-26}\\\frac{K\times q_1\times q_2}{U_{\infty}} = 2.15\times 10^{-14}[/tex]
[tex]r_0^2 - 2.15\times 10^(-14)\times r_0 - 1.69\times 10^{-24} = 0;
r_0 = 1.31\times 10^{-12} m[/tex]
b) [tex]r_0^2 - 2.15\times 10^{-14}\times r_0 - 1.21\times 10^{-26} = 0
\\r_0 = 1.21\times 10^{-13} m
[/tex]
c) [tex]r_0^2 - 2.15\times 10^{-14}\times r_0 - 1.00\times 10^{-28} = 0
\\r_0 = 2.54\times 10^{-14} m[/tex]
