Question

Consider the following.

n + ²³⁵U → ⁹⁰Kr + ¹⁴⁴Ba + 2 n

(a) Calculate the energy (in MeV) released in the neutron-induced fission reaction above, given m(⁹⁰Kr) = 89.919524 u and m(¹⁴⁴Ba) = 143.922941 u. (Assume 1 u = 931.5 MeV/c².) MeV

(b) Confirm that the total number of nucleons and total charge are conserved in this reaction.

A_i = A_f=

Q_i = Q_f=

Answer 1

a.)

Mass defect is given by:

dm = Mass of neutron + mass of 235U - (mass of 90Kr + Mass of 144Ba + 2*Mass of neutron)

Given that

Mass of 1 neutron = 1.008664916 u

mass of 235U = 235.043929918 u

mass of 90Kr = 89.919524 u

Mass of 144Ba = 143.922941 u

So,

dm = 1.008664916  + 235.043929918 - (89.919524 + 143.922941 + 2*1.008664916)

dm = mass defect = 0.1928 u

Energy released is given by:

E = dm*c^2

c^2 = 931.5 Mev/u

So,

E = (0.1928 u)*(931.5 MeV/u) = 0.1928*931.5

E = energy released in fission reaction = 179.5932 MeV = 179.6 MeV

B.

Given reaction is:

Number of nucleons in neutron = 1

Number of nucleons in U = 235

Number of nucleons in Kr = 90

Number of nucleons in Ba = 144

So

total nucleons on the left side = 1 + 235 = 236

total nucleons on the right side = 90 + 144 + 2 = 236

Now number of protons in neutron = 0

Number of proton in U = 92

Number of proton in Kr = 36

Number of proton in Ba = 56

total charge on the left side = 0 + 92 = 92

total charge on the right side = 36 + 56 = 92

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