Nuclear Physics(Semi Empirical Mass Formula)

[Catty]

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...

 

[mfb]

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)).

 

[Catty]

thank you.