Collision of Alpha particle with a Gold Nucleous

[lebprince]

Homework Statement

An alpha particle (a helium nucleus, containing 2 protons and 2 neutrons) starts out with kinetic energy of 9.5 MeV (9.5106 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.

a) What is the initial momentum of the alpha particle? (You may assume its speed is small compared to the speed of light). I got the right answer here <1.427e-19,0,0>

b) What is the initial momentum of the gold nucleus? I also got this right <0,0,0>

c) What is the final momentum of the alpha particle, long after it interacts with the gold nucleus? Am stuck on this part.

The Attempt at a Solution

i was able to solve parts a and b. for part C i tried getting 2 equations so i got Kf1+Kf2 = 1.52e-12 which is the given Kinetic energy from the problem changed to Joules, and i got P1f + P2f = 1.47e-19 , and i don't know what else do if am on the right track. any help would be appriciated thanks.

 

[lebprince]

Ok there is a question after part c asking me for the final momentum of the gold nucleus long after the interaction with the alpha particle. i got the answer for that which was 2 * p1i, p1i is the initial momentum of the alpha particle. so 2 * 1.427e-19. now for part C i tried using the momentum principle and set p1i + p2i = p1f + p2f since i have p1i and p2i and p2f i can find part c which p1f but am not getting the right answer any help?

 

[thomate1]

Answer:

To solve part c, you can use the conservation of momentum and energy equations. Since the alpha particle and the gold nucleus are the only objects involved in the collision, the total initial momentum and energy must be equal to the total final momentum and energy. This can be expressed as:

Initial momentum (alpha particle) + Initial momentum (gold nucleus) = Final momentum (alpha particle) + Final momentum (gold nucleus)

and

Initial kinetic energy (alpha particle) + Initial kinetic energy (gold nucleus) = Final kinetic energy (alpha particle) + Final kinetic energy (gold nucleus)

Using the values from parts a and b, we can write the equations as:

(1.427e-19,0,0) + (0,0,0) = (P1f, P2f, 0) + (0,0,0)

and

(9.5 MeV, 0, 0) + (0, 0, 0) = (K1f, K2f, 0) + (0, 0, 0)

From the first equation, we can see that P1f must be equal to 1.427e-19, and since the gold nucleus is initially at rest, P2f must be equal to 0.

Substituting these values into the second equation, we can solve for the final kinetic energy of the alpha particle, K1f:

(9.5 MeV, 0, 0) = (K1f, 0, 0)

Therefore, K1f = 9.5 MeV, which is the same as the initial kinetic energy of the alpha particle. This means that after the collision, the alpha particle will still have the same kinetic energy and will continue moving in the +x direction with the same momentum as before.

In summary, the final momentum of the alpha particle is (1.427e-19, 0, 0) and its final kinetic energy is 9.5 MeV.