Quantum tunneling and radioactive decay.


by bayan
Tags: decay, quantum, radioactive, tunneling
bayan
bayan is offline
#1
May25-12, 03:21 AM
P: 202
1. The problem statement, all variables and given/known data

The edge of a nucleus can be roughly modeled as a square potential barrier. An alpha particle in an unstable nucleus can be modeled as a particle with a specific energy, bouncing back and forth between these square potential barrier.

Consider a nucleus of radius r and an alpha particle with kinetic energy E (i.e., let the potential energy within the nucleus be zero) and mass m.

Assuming that the alpha particle moves along a diameter of the nucleus and that it moves at low enough speed that relativistic effects are negligible, what is the time tau between successive encounters between each edge of the nucleus and the alpha particle?

Express your answer in terms of [itex]K_{e}[/itex], [itex]r[/itex], and [itex]m[/itex].

2. Relevant equations

[itex]K_{e}=mv^{2}[/itex]



3. The attempt at a solution

[itex]v^{2}=\frac{(2r)^{2}}{t^{2}}[/itex]

[itex]\frac{K_{e}}{m}=\frac{(2r)^{2}}{t^2}[/itex]

[itex]\sqrt{\frac{K_{e}}{m}}=\frac{2r}{t}[/itex]

[itex]2r\sqrt{\frac{m}{K_{e}}}={t}[/itex]

Have I used the right approach to this problem? and have I got the correct answer?

Thanks in advance
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bayan
bayan is offline
#2
May25-12, 07:55 PM
P: 202
Seems like the answer does not need the 2 in my answer, which means I have made a mistake in my approach or calculation. Anyone able to help out?

Thanks.
Curious3141
Curious3141 is offline
#3
May25-12, 08:50 PM
HW Helper
P: 2,874
I'm not really sure about your approach, but there is a factor of half missing in your KE equation:

[tex]K_e = \frac{1}{2}mv^2[/tex]

Does that help?

bayan
bayan is offline
#4
May26-12, 01:07 AM
P: 202

Quantum tunneling and radioactive decay.


Quote Quote by Curious3141 View Post
I'm not really sure about your approach, but there is a factor of half missing in your KE equation:

[tex]K_e = \frac{1}{2}mv^2[/tex]

Does that help?
I should hope so, lol

Thanks :)


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