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 9.9 MeV (9.9 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 = (1.45 middot 10^-19, 0, 0) kg middot m/s What is the initial momentum of the gold nucleus? P_Au, i = (0, 0, 0) kg middot m/s What is the final momentum of the alpha particle, long after it interacts with the gold nucleus? P_alpha, f = (1.45 middot 10^-19, 0, 0) kg middot m/s What is the final momentum of the gold nucleus, long after it interacts with the alpha particle? P_Au, f = (2.9 middot 10^-19, 0, 0) kg middot m/s What is the final kinetic energy of the alpha particle? K_alpha, f = -1.45e-19 J What is the final kinetic energy of the gold nucleus? K_Au, f = 1.287e-13 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 = 2.4e-14 m

Explanation / Answer

using conservation of momentum

initial momentum of alpha particle + initial momentum of gold nucleus = final momentum of alpha particle + final momentum of gold nucleus

((1.45 x 10-19) i + 0 j + 0 k) + (0 i + 0 j + 0 k) = final momentum of alpha particle + ((2.9 x 10-19) i + 0 j + 0 k)

final momentum of alpha particle = (- (1.45 x 10-19) i + 0 j + 0 k)

Palphafinal = final momentum of alpha particle = (- (1.45 x 10-19) i + 0 j + 0 k)

|Palphafinal | = magnitude of final momentum of alpha particle = 1.45 x 10-19

malpha = mass of alpha particle = 6.64 x 10-27 kg

Kinetic energy is given as

KE = P2/(2m)

hence ,

KEalphafinal = final KE of alpha particle = (|Palphafinal |)2/(2 malpha ) = (1.45 x 10-19)2 /(2 x 6.64 x 10-27)

KEalphafinal = 1.583 x 10-12 J

Pgoldfinal = final momentum of gold particle = ( (2.9 x 10-19) i + 0 j + 0 k)

|Pgoldfinal | = magnitude of final momentum of gold particle = 2.9 x 10-19

mgold = mass of gold particle = 79 x 1.67 x 10-27 + 118 x 1.67 x 10-27 = 328.99 x 10-27 kg

Kinetic energy is given as

KE = P2/(2m)

hence ,

KEgoldfinal = final KE of gold particle = (|Pgoldfinal |)2/(2 mgold ) = (2.9 x 10-19)2 /(2 x 328.99 x 10-27 )

KEgoldfinal = 1.278 x 10-13 J

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.