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## Main Question or Discussion Point

1.

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

(i). Δm = m(reactants) - m(products)

(ii). E = mc^2

(iii). u = atomic mass unit

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

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......

**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......

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