Minimum Electric Potential Difference Required to Bring Alpha Particle to Rest

[Mspike6]

Hey..

i got this question in a unit assessment test... i solved it (or atleast i think so :P ) but the answer doesn;t make sense . so i want someone to tell me if i did something wrong.

A plutonium-239, is initialy at rest, undergoes an alpha decay \, to produce a uranium 235 nucleus. the uranium-235 nucleus has a mass of 3.90*10^-25 Kg, and moves away from the location of the decay at speed of 2.62 *10⁵ m/s


Determine the minimum electric potential difference that is required to bring the alpha particle to rest.
Marks will be awarded based on the physics principles you provide, the formulas you state, the substitutions you show, and your final answer. (12 marks)


Solution:
To bring the alpha particle to rest, we must apply Electric Force equal to the alpha particle’s Kinetic energy.

qE= ½ mv2 –(1)

V=E/q
E= Vq –(2)

Substitute (2) into (1)

Vq2 = ½mv2
V= ½ mv2/q2

m=6.65*10^-27Kg
q= 3.2*10^-19 C

V= 2.3*10²¹ V

 

[cepheid]

Solution:
To bring the alpha particle to rest, we must apply Electric Force equal to the alpha particle’s Kinetic energy.

No, here is the error. You can't equate two *different* physical quantities to each other. Saying that 'force = energy' is just as nonsensical as saying that 'time = distance.'

What must actually be true is that all of the particle's kinetic energy is transformed into electric potential energy. I.e. Initial kinetic energy = final potential energy. This problem is about conservation of energy.

Equivalently, if you were to say:

*Work done* by electric force = change in kinetic energy,

that would be fine. In fact, the work done and the change in potential energy will just differ by a minus sign. Saying that work = energy is fine, because these are the *same* type of physical quantity (measured using the same units).

 

[Mspike6]

Ya, what you say makes perfect sense, i don't know how did i do sucha mistake..

okay let's say that i got Delta E, which is Work done, But then where do i go ? can i use V=E/Q to get V ?

Thanks a lot mate, really appreciated. :P

 

[cepheid]

Ya, what you say makes perfect sense, i don't know how did i do sucha mistake..

okay let's say that i got Delta E, which is Work done, But then where do i go ? can i use V=E/Q to get V ?

Thanks a lot mate, really appreciated. :P

Hmm, yeah, basically, I think that's what you do. You need to know the charge of an alpha particle.

 

[Mspike6]

Ok, i solved it but i hope i got it right this time!

First: Conservation of momentum.

Mv initial = mv Duaghter pa. +mv Alpha pa.

mv Duaghter pa. =-mv Alpha pa

(3.90*10^-25)(-2.62 *10⁵) = the momentum of the duaghter particle...the negative on the speed indicate that it was moving in the opposite direction.

=-1.02*10^-19

mv alpha particle = 1.02*10^-19

V= 1.02*10^-19 / 6.65*10^-27 = 1.53*10⁷

1/2(m)(v²) = 7.78*10^-13

Delta E = 1/2 mv² (Since it will bring it to stop)
Delta E= 7.78*10^-13
V=Delta E /q

V= 7.78*10^-13 / (2(1.60*10^-19))

V=2.43*10⁶



I really hope this is correct :D

Thanks guys