Beta minus decay?

  • Thread starter hasnain721
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  • #1
hasnain721
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[SOLVED] beta minus decay?

Hi,
Here is a question from beta minus decay which i could not understand :


http://img360.imageshack.us/img360/2512/59060072re9.jpg [Broken]


I could do the first bit but not the second. I believe that it has something to do with neutrinos and missing energy but i am not 100% sure about it.


Thanks.
 
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Answers and Replies

  • #2
astrorob
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hasnain,

Beta decay involves the transition of a parent nucleus into a daughter nucleus with two sideproducts; an electron and an antineutrino.

Since the nucleus is so massive, it carries little kinetic energy away from the reaction (it recoils very little), the vast majority of the energy goes into the two sideproducts.

Now let me ask you a question:

If there was only the electron as a sideproduct, would there be a continuous range of energies as we see here?
 
  • #3
hasnain721
40
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hasnain,

Beta decay involves the transition of a parent nucleus into a daughter nucleus with two sideproducts; an electron and an antineutrino.

Since the nucleus is so massive, it carries little kinetic energy away from the reaction (it recoils very little), the vast majority of the energy goes into the two sideproducts.

Now let me ask you a question:

If there was only the electron as a sideproduct, would there be a continuous range of energies as we see here?

hi,
Thanks a lot for replyin astrorob.


As far as the question goes, i believe that if only an electron was a side product them the graph would have been a straight line parallel to the y axis. Is that right?
 
  • #4
astrorob
140
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hasnain,

You are correct. There would be only a small range of energies the electron could carry away and this would manifest itself (essentially) as a straight line at x=0.78.
 
  • #5
hasnain721
40
0
hasnain,

You are correct. There would be only a small range of energies the electron could carry away and this would manifest itself (essentially) as a straight line at x=0.78.



Thanks a lot astrorob.

CHeers!
 
  • #6
astrorob
140
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You're welcome.
 

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