An alpha particle with kinetic energy 12.5 MeV makes a collision with lead nucle

Question

An alpha particle with kinetic energy 12.5 MeV makes a collision with lead nucleus, but it is not "aimed" at the center of the lead nucleus, and has an initial nonzero angular momentum (with respect to the stationary lead nucleus) of magnitude L = p_0b, where p_0 is the magnitude of the initial momentum of the alpha particle and 6=1.50 times 10^-12 m. (Assume that the lead nucleus remains stationary and that it may be treated as a point charge. The atomic number of lead is 82. The alpha particle is a helium nucleus, with atomic number 2.) Repeat for 6=1.20 times 10^-13 m. Repeat for n= 1.00 times 10^-14 m

Explanation / Answer

here,
Assuming
v0 be the velocity
r0 the distance at the closest approach.

the initial kinetic energy U() = 12.5 MeV = 2.0027*10^-12 J
charge on alpha particle, q1 = 3.2*10^-19 C
Charge on lead nucleus, q2 = 131.2*10^-19 C

The total energy at closest approach = kinetic energy + potential energy
The total energy there is then :
PE = K*q1*q2/r0
KE = 0.5*m*v0^2
U(r0) = K*q1*q2/r0 + 0.5*m*v0^2 ----------------(1)

Using conservation of angular momentum,
initial = Final
v*m*b = v0*m*r0

v0 = v*b/r0 ---------------------------(2)

Using 2 in 1 we get :
U(r0) = K*q1*q2/r0 + 0.5*m*v^2*b^2/r0^2

this must equal Intital KE
U = (K*q1*q2)/r0 + (0.5*m*v^2*b^2)/r0^2

also, 0.5*m*v^2 = U

U = K*q1*q2/r0 + U*b^2/r0^2
U*r0^2 - (K*q1*q2)*r0 - U*b^2 = 0
r0^2 - (K*q1*q2/U)*r0 - b^2 = 0
r0^2 - (K*q1*q2/U)*r0 - b^2 = 0 -------------------------(3)

Case 2 : when b = 1.20*10^-13 m , equation3 can be written as :
r0^2 - ( (9*10^9 * 3.2*10^-19 * 131.2*10^-19) / (2.0027*10^-12) )*r0 - (1.20*10^-13 )^2 = 0

solving quadratic equation we get :
r0 = 1.298 * 10^-13 m

Case 3 : when b = 1.00*10^-14 m , equation3 can be written as :
r0^2 - ( (9*10^9 * 3.2*10^-19 * 131.2*10^-19) / (2.0027*10^-12) )*r0 - (1.00*10^-14)^2 = 0

solving quadratic equation we get :
r0 = 2.318 * 10^-14 m

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