An alpha particle with kinetic energy 11.0 MeV makes a collision with lead nucle

Question

An alpha particle with kinetic energy 11.0 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 = p_0 b, where p_0 is the magnitude of the initial momentum of the alpha particle and b = 1.40 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.) What is the distance of closest approach? r = m Repeat for b = 1.10 times 10^-13 m. r = m Repeat for b = 1.00 times 10^-14 m. r = m

Explanation / Answer

angular momentum and energy must be conserved

At the distance of closest approach the speed is not zero .

E = K+U

q1 = 2e

q2 = 82e

mv1*b = mv2*r

E1 = E2

E1 = mv2^2/2 + kq1q2/r

E1 = 11 MeV = 1.76 x 10^-12 J

v2 = v1*b/r

E1 = E1*b^2/r^2 + kq1q2/r

(E1)r^2 - (kq1q2)r -E1*b^2 = 0

b = 1.4 x 10^-12

1.76 x 10^-12 * r^2 - 3.77856 x10^-26r - 3.4496 x 10^-36 = 0

by solving this quadratic equation

r = 1.41 x 10^-12 m

Part b

b = 1.1 x 10^-13 m

1.76 x 10^-12 * r^2 - 3.77856 x10^-26r - 2.1296 x 10^-38 = 0

r = 1.21 x 10^-13 m

part c )

b = 1 x 10^-14

1.76 x 10^-12 * r^2 - 3.77856 x10^-26r - 1.76x10^-40 = 0

r = 2.54 x 10^-14 m

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