Please explain the question 3 part a, b, and c M(^238_92 U) = 236.045562u m(^230

Question

Please explain the question 3 part a, b, and c

M(^238_92 U) = 236.045562u m(^230_90 U) = 234.043593u m(^234_92 U) = 234.040946u m(^238_94 Pu) = 238.04955u m(^234_91 Pa) = 234.043300u m(^234_93 Np) = 234.046560u Identity the unknown isotopes X and Y in the following decay ^238_94 Pu rightarrow X + alpha X rightarrow Y + beta^+ X = ^2345_92 u Y = ^234_91 Pa Calculate the energy released in the alpha-decay oi ^238_94 Pu. Q_alpha = (238.049555 - 234.0433 - 4,oo2602)u times 9315 = 5.6 Mev Calculate the binding energy per nucleon of ^238_94 Pu. B = (94 times 1.007825 + 144 times 1.oo8665 - 238.049555) times 931.5 B = 1.8 times 10^3 Mev B/A = 76 Mev

Explanation / Answer

(a)
Alpha particles consist of two protons and two neutrons , Can be writteen as, 4H2

Now we see, 238U94 ==> X +  4H2

Possible Value of X, = 234U92

Similarly, + is a positron emission,
234U92 ==> Y + 01
Possible Value of Y, = 234PA91

(b)
Mass of reactants,
238U94 = 238.049555 u

Mass of Products,
234U92 = 234.040946 u
4H2 = 4.001506 u

Energy Released = Difference in mass
1 u = 931.5 Mev

Q = (238.049555 - 234.040946 - 4.001506) * 931.5 Mev
Q = 6.62 Mev

(c)
The difference between the mass of a nucleus and the sum of the masses of the nucleons of which it is composed is called the mass defect. and when it is converted to energy is called Binding Energy.
The mass of a proton is 1.00728 amu and a neutron is 1.00867 amu.
No of proton = 94
No of neutron = 238 - 94 = 144

Now Mass Defect = (94 * 1.00728 + 144 * 1.00867 - 238.049555) u
Mass Defect = 1.883245 u

Binding Energy = 1.883245 * 931.5 Mev
Binding Energy = 1754.24 Mev
Binding Energy / nucleon = 1754.24 / 238
Binding Energy / nucleon = 7.37 Mev

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