Question

An alpha particle with kinetic energy 14.0 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=p0b, where p0 is the magnitude of the initial momentum of the alpha particle and b=1.50×10−12 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.)

A) What is the distance of closest approach?

B) Repeat for b=1.50×10⁻¹³ m .

C) Repeat for b=1.50×10⁻¹⁴ m .

Answer 1

Answer :

The total energy at closest approach is the kinetic energy + potential energy at that point. Let v0 be the velocity and r0 the distance at the closest approach. The total energy there is then

PE = K*q1*q2/r0
KE = 0.5*m*v0²

Total energy U(r0) = K*q1*q2/r0 + 0.5*m*v0²

Using conservation of angular momentum,

initial = v∞*m*b
at r0 = v0*m*r0

these must be equal so v∞*m*b = v0*m*r0

v0 = v∞*b/r0

substitute this for v0 in the energy equation

U(r0) = K*q1*q2/r0 + 0.5*m*v∞²*b²/r0²

this must equal the initial kinetic energy U(∞) (14 MeV)

U∞ = K*q1*q2/r0 + 0.5*m*v∞²*b²/r0²

however, 0.5*m*v∞² = U∞

U∞ = K*q1*q2/r0 + U∞*b²/r0²

U∞*r0² - (K*q1*q2)*r0 - U∞*b² = 0

r0² - (K*q1*q2/U∞)*r0 - b² = 0 ........................(1)

U∞ = 2.24*10^-12 J

q1 to be 3.204 * 10^-19 and q2 as 1.314 * 10 ^-17
K*q1*q2 = 3.789*10^-26
K*q1*q2/U∞ = 1.689*10^-38
b=1.50×10−12 m

r0² - 1.689*10^-38*r0 - 2.25*10^-24 = 0
r0 = 1.5e-12

(b) b=1.50×10−13 m

r0² - 1.689*10^-38*r0 - 2.25*10^-26 = 0
r0 = 1.5e-13

(c) b=1.50×10−14 m

r0² - 1.689*10^-38*r0 - 2.25*10^-28 = 0
r0 = 1.5e-14