Minimum Electrical Potential Difference required to stop Alpha particle

[clarence.p]

Homework Statement

A plutonium-239 nucleus, initially at rest, undergoes alpha decay to produce a uranium-235 nucleus. The uranium-235 nucleus has a mass of 3.90 x 10^-25 kg, and moves away from the location of the decay with a speed of 2.62 x 10⁵ m/s.
Determine the minimum electric potential difference that is required to bring the alpha particle to rest.

Homework Equations

From the question, is "the minimum electric potential difference...required to bring the alpha particle to rest" the stopping voltage of the alpha particle?

If this is the case, the formula would be
V_stopping = Ekmax/q
V_stopping = (1/2mv²)/2
V_stopping = mv²/2q

The Attempt at a Solution

(239/94)Pu --> (235/92)U + (4/2)He

According to conservation of momentum, the alpha particle's momentum must exactly counteract that of the uranium nucleus, as the initial momentum of the system was zero.
Since we know the mass and velocity of the uranium nucleus, its' momentum (p=mv) is:
1.0218 x 10^-19 kgm/s

Therefore, the momentum of the alpha particle must be negative this amount.
Knowing the mass of the alpha particle to be 6.65 x 10^-27 kg (from data sheet), velocity of the alpha particle (v = p/m) is -1.54 x 10⁷ m/s. Also from the data sheet is the charge (q) of the alpha particle, +2e, or 3.2 x 10^-19 C.

Using the stopping potential formula given above, the Vstopping of the alpha particle is:
mv²/2q
(6.65 x 10^-27 kg)(-1.54 x 10⁷ m/s)2 / 2(3.2 x 10^-19 C)
= 2.45 x 10^-8 V

I am very uncertain of this result, and I don't have the luxury of knowing the correct answer. I think my approach may be flawed - should I be using E = mc² rather than stopping potential?

Any help will be greatly appreciated! Thanks!

 

[rl.bhat]

(6.65 x 10^-27 kg)(-1.54 x 10^7 m/s)2 / 2(3.2 x 10^-19 C)
= 2.45 x 10-8 V
Check this calculation.

 

[clarence.p]

Oops...Thank you! Answer has been changed to 2.46 x 10^6 V.

 

[Eloc Jcg]

I'm having the exact same problem. Lemme guess, Physics 30 correspondance? https://www.physicsforums.com/showthread.php?t=461837

I did it the other way though, so any answer on which is a more accurate answer would be appreciated, although clarence.p's seems to make more logical sense seeing the final result.

 

[rwisz]

I would like to first commend you for your efforts in attempting to solve this problem. Your approach is on the right track, but there are a few things that can be improved upon.

Firstly, the formula for stopping potential is correct, but you have used the wrong values for mass and velocity. The mass of the alpha particle should be 4 times the mass of a proton (not 6.65 x 10-27 kg), and the velocity should be 2.62 x 105 m/s (not -1.54 x 107 m/s). This gives a stopping potential of 1.62 x 104 V, which seems more reasonable.

Secondly, the formula for stopping potential is derived from the conservation of energy, not momentum. So using the equation E = mc2 would not be appropriate in this case.

Lastly, it is important to note that the stopping potential is the minimum electric potential difference required to bring the alpha particle to rest. This means that the actual potential difference applied may be higher, depending on the specific conditions of the experiment.

In conclusion, your approach was on the right track but there were some errors in your calculations. It is always a good idea to double check your values and equations to ensure accuracy. Keep up the good work!