Nuclear Physics(Semi Empirical Mass Formula)

  • Thread starter Catty
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  • #1
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Main Question or Discussion Point

1. Question : Use the Semi Empirical Mass Formula to estimate the energy released in the spontaneous fission reaction;

235/92 Uranium -----> 87/35 Bromine + 145/57 Lanthanum + 3n




2. Equations used :

(i). Δm = m(reactants) - m(products)
(ii). E = mc^2
(iii). u = atomic mass unit



3. Attempt to solution:

atomic mass for 235/92 U = 235.043929 u
" " " 87/35 Br = 86.920711 u
" " " 145/57 La = 144.921765 u
" " " 1/0 n = 1.009 u

thus, Δm = m(reactants) - m(products)
= 235.043929 - (86.920711 + 144.921765 + (3*1.009))
= 2.19245u

apply E = mc^2 = 2.19245u * (931.5 Mev/c^2)/u = 2043MeV




4. Problem:

i'm not sure if i'm answering the question right since it says "use the SEMF" , but this is how i am calculating the energy so far. Is this how i use the SEMF to calculate the energy, if not, how should the SEMF appear in the calculations? please kindly help......
 
Last edited:

Answers and Replies

  • #2
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I think you are supposed to estimate the nucleon masses with the SEMF instead of looking them up.

I get a different result for 235.043929 - (86.920711 + 144.921765 + (3*1.009)).
 
  • #3
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thank you.
 

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