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The Brogile wavelength of an Alpha particle

  • Thread starter mrausum
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  • #1
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Homework Statement



An alpha particle of energy 5.78 MeV is emitted from a particular nucleus.
Calculate the de Broglie wavelength of the alpha particle. How does it compare
with the nuclear diameter, which is known to be approximately 2x10−14 m.


Homework Equations



E=h/p and E=hf

The Attempt at a Solution



I was thinking of using E=hc/lamda but this would be wrong because alpha particles don't travel at c.
 

Answers and Replies

  • #2
rock.freak667
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The energy of 5.78MeV is entirely kinetic.
 
  • #3
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so how do i do the question?
 
  • #4
rock.freak667
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  • #5
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well it hasn't given the mass. But are you saying work out v from E=1/2mv^2, then f from E/h. Then wavelength from lamda=v/f?
 
  • #6
rock.freak667
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well it hasn't given the mass. But are you saying work out v from E=1/2mv^2, then f from E/h. Then wavelength from lamda=v/f?
find v, then use

[tex]\lambda = \frac{h}{mv}[/tex]

I am sure you can look up the mass of an alpha particle. I am not sure how else you can solve this problem without my suggestion though.
 
  • #7
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how do you get lambda = h/mv
i know E = hc/lambda = h/mv
so, if i combine both equations, shouldnt i suppose to get lambda = cmv ?
did i miss something or what.
 
  • #8
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how do you get lambda = h/mv
i know E = hc/lambda = h/mv
so, if i combine both equations, shouldnt i suppose to get lambda = cmv ?
did i miss something or what.
p = h/lamda
lamda = h/p=h/mv

You can't use E = hc/lamda because alpha particles don't travel at the speed of light.
 

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