An alpha particle (a helium nucleus, containing 2 protons and 2 neutrons) starts

Question

An alpha particle (a helium nucleus, containing 2 protons and 2 neutrons) starts out with kinetic energy of 9.6 MeV (9.6 times 10^6 eV), and heads in the +x direction straight toward a gold nucleus (containing 79 protons and 118 neutrons). The particles are initially far apart, and the gold nucleus is initially at rest. Answer the following questions about the collision. What is the initial momentum of the alpha particle? (You may assume its speed is small compared to the speed of light). p_alpha, i = kg m/s What is the initial momentum of the gold nucleus? p_Au, i = kg m/s What is the final momentum of the alpha particle, long after it interacts with the gold nucleus? p_alpha, f = kg m/s What is the final momentum of the gold nucleus, long after it interacts with the alpha particle? p_Au, f = kg m/s What is the final kinetic energy of the alpha particle? K_alpha, f = J What is the final kinetic energy of the gold nucleus? K_Au, f = J Assuming that the movement of the gold nucleus is negligible, calculate how close the alpha particle will get to the gold nucleus in this head-on collision. r_closest = m

Explanation / Answer

Mass of alpha particle, m1 = 4 x1.673 x 10^-27 kg = 6.692 x 10^-27 kg

Energy = 9.6 MeV = 9.6 x 10^6 x 1.6 x 10^-19 J = 1.536 x 10^-12 J

v1i = 2.143 x 10^7 m/s

p1i = sqrt(2m K ) = 1.43 x 10^-19 kg m/s

P1i = < 1.43e-19, 0, 0 > ...........Ans

P2i = 0

for elastic collision,

velocity of approach = velocity of separation

2.143 x 10^7 = v1 + v2

v2 = 2.143 x10^7 - v1

suppose after collision speed of alpha particle is v1 in theopposite direction of gold is v2 in the same direction.

4u x 2.143 x 10^7 + 0 = 4u(-v1) + 197v2

197(2.143 x 10^7 - v1) - 4v1 = 8.57 x 10^7

v1 = 2.058 x 10^7 m/s

final momentum of alpha particle, magnitude = m1 v1 = 1.377 x 10^-19 kg m./s

Ans: < -1.377e-19, 0, 0> ..........Ans

v2 = 8.5 x 10^5 m/s

momentum = 197 x 1.673 x 10^-27 x 8.5 x 10^5 = 2.801 x 10^-19 kg m/s

Momentum of gold nucleus = <2.801e-19, 0, 0> kg m/s ........Ans

Ki = m1 v1i^2 / 2

= 1.537 x 10^-12 J ...........Ans

Kf = m1 v1^2 /2 + m2 v2^2 / 2

= 1.537 x 10^-12 J .......Ans

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