Suggested+Solutions+to+Mass,+Binding+Energy+and+the+Liquid+Drop+Model

Exercises with Mass, Binding Energy and the Liquid Drop Model > M(n) = 1.008665 u > M( 1 H) = 1.007825 u > M( 4 He) = 4.002603 u > M( 56 Fe) = 55.934938 u > M( 142 Ce) = 141.909197 u > M( 238 U) = 238.050788 u
 * 1:**
 * 1) Mass excess is given by ΔM(Z,A) = M(Z,A)-A. The masses can therefore be found with M(Z,A) =ΔM (Z,A ) + A by using 1 u = 931.5 MeV we get:
 * 1) The most stable nuclei is 56 Fe. It has the highest binding energy compared to the number of nucleons.
 * 2) 1.00 kg 2 H is 496.5 moles by fusion 248.2 mole 4 He is created which is 993.6 g. The difference in mass is equal to the energy 1000 -993.6 g = 6.36 g. The energy is then 5.7•10 14 J, 3.56•10 27 MeV or 1.5910 8 kWh.
 * 3) 1.00 kg 233 U fission to 92 Rb, 138 Cs and three neutrons. The mass difference is 0.785 g. Which gives 7.1•10 13 J, 4.40 •10 26 MeV or 1.96•10 7 kWh.
 * 4) The energy from the fission is distributed in different ways: most of it goes to the kinetic energy of for the fission products. Other parts go to the kinetic energy of the neutrinos and the neutrons, and some of it goes to “prompt” gamma-rays and beta/gamma rays from the fission products.


 * 2: **

Binding energy B per nucleon:

math \frac{B}{A}=\frac{Z\cdot M(^{1}H)+N\cdot M(^{1}n)-M(A,Z)}{A}=\frac{12 \cdot 1.00782503207 + 12 \cdot 1.0086649156 - 23.985041}{24}=0.00868 \, u math

It is 0.00868 u per nucleon and 8.26 MeV per nucleon


 * 3: **

math \frac{E_{nucleus}}{E_{electron}}=\frac{8111.493 \, keV}{5.14\cdot 10^{-3} \, keV}=1.6\cdot 10^{5}

math

1.6•10 5 times larger binding energy for the nucleus.


 * 4: **
 * 1) m n  = 1.67 •10 -27 kg
 * 2) m e  = 9.11 •10 -31 kg
 * 3) m u  = 1.66•10 -27 kg


 * 5: **
 * 1) <span style="font-family: 'Times New Roman','serif'; font-size: 16px;">8.55 MeV/nucleon.
 * 2) <span style="font-family: 'Times New Roman','serif'; font-size: 16px;">8.79 MeV/nucleon.
 * 3) <span style="font-family: 'Times New Roman','serif'; font-size: 16px;">7.86 MeV/nucleon.

<span style="font-family: 'Times New Roman','serif'; font-size: 16px;">The energy <span style="font-family: 'Times New Roman','serif'; font-size: 16px;">M( 235 <span style="font-family: 'Times New Roman','serif'; font-size: 16px;">U) - (M( 131 <span style="font-family: 'Times New Roman','serif'; font-size: 16px;">Xe + M( 101 <span style="font-family: 'Times New Roman','serif'; font-size: 16px;">Ru) + 3M(n)) = <span style="font-family: 'Times New Roman','serif'; font-size: 16px;">(ΔM( 235 <span style="font-family: 'Times New Roman','serif'; font-size: 16px;">U) + 235) - (Δ( 131 <span style="font-family: 'Times New Roman','serif'; font-size: 16px;">Xe) + 131) - (ΔM( 101 <span style="font-family: 'Times New Roman','serif'; font-size: 16px;">Ru) + 101) - 3(ΔM(m) + 1) <span style="font-family: 'Times New Roman','serif'; font-size: 16px;">= ΔM( 235 <span style="font-family: 'Times New Roman','serif'; font-size: 16px;">U) - ΔM( 131 <span style="font-family: 'Times New Roman','serif'; font-size: 16px;">Xe) - ΔM( 101 <span style="font-family: 'Times New Roman','serif'; font-size: 16px;">Ru) - 3ΔM(n) <span style="font-family: 'Times New Roman','serif'; font-size: 16px;">=40.916 + 88.421 + 87.952 - 3• 8.071 = 193.08 MeV
 * <span style="font-family: 'Times New Roman','serif'; font-size: 16px;">6: **

<span style="font-family: 'Times New Roman','serif'; font-size: 16px;">Energy usage per 10 km: <span style="font-family: 'Times New Roman','serif'; font-size: 16px;">1 L = 700 g. Mm(C8H18) = 114 g/mole, this means that 700 g is 6.14 mol. <span style="font-family: 'Times New Roman','serif'; font-size: 16px;">6.14 mole•5500 kJ/mole = 33771.93 kJ, with a fuel efficiency of 18% the usage per 10 km is 6078.94 kJ/(10 km). <span style="font-family: 'Times New Roman','serif'; font-size: 16px;">1 g 235 <span style="font-family: 'Times New Roman','serif'; font-size: 16px;">U=4.25•10 -3 <span style="font-family: 'Times New Roman','serif'; font-size: 16px;"> mole = 2.56•10 21 <span style="font-family: 'Times New Roman','serif'; font-size: 16px;"> atoms. <span style="font-family: 'Times New Roman','serif'; font-size: 16px;">If all of the atoms fission and assuming 100% efficiency this will give 5.12•10 29 <span style="font-family: 'Times New Roman','serif'; font-size: 16px;"> eV = 8.22•10 7 <span style="font-family: 'Times New Roman','serif'; font-size: 16px;">kJ Which will last for 135190 km.
 * <span style="font-family: 'Times New Roman','serif'; font-size: 16px;">7: **

ΔG = ΔGproduct - ΔGreactant= - 237 kJ/mol. Which means that for each H 2 O molecule it is generated 237kJ/NA = 2.456•10-6 MeV. Comparing to 0.0303 u = 23.2 MeV released by fusion to helium. We see that the nuclear reactions functions on a scale 10 6 times bigger than chemica<span style="font-family: 'Times New Roman','serif'; font-size: 16px;">l.
 * <span style="font-family: 'Times New Roman','serif'; font-size: 16px;">8: **

Fusion of 2 H gives 5.75•10 11 J/gram approximately ten times more energy then uranium which gives 8.2•10 10 J/gram.
 * <span style="font-family: 'Times New Roman','serif'; font-size: 16px;">9: **

In this problem it is useful to keep in mind the binding energy equation: math B_{tot}(A,Z)=a_{v}A-a_{s}A^{2/3}-a_{c}\frac{Z^{2}}{A^{1/3}}-a_{a}\frac{A-2Z}{A}\pm\delta math
 * <span style="font-family: 'Times New Roman','serif'; font-size: 16px;">10: **

Most isobars (A= constant, Z varies) will have binding energy when plotted as a function of Z creating a parable. The last part of the equation in the formula for binding energy delta expresses the energy gained when the proton and/or the neutron are in a couple. A nucleus with an even amount of both will have a positive δ, higher binding energy, an unpaired nucleus either proton or neutron will have a δ of zero, a nucleus with a unpaired neutron and proton will have a negative δ. Therefore for odd-nuclei the binding energy is independent of which nucleon is unpaired. The binding energy as a function of z will have a minimum value for a certain stable nucleus. For even nuclei with a given A value they will have an even-even, or an odd-odd configuration when Z varies. This will create two curves where the odd-odd nuclei will have a lower binding energy than the even-even nuclei. Odd-odd nuclei will therefore have more possibilities when decaying to a level with higher binding energy, either with a β - decay or a β + decay. Both of these mechanisms will lead to a more stable nucleus.

11: <span style="font-family: 'Times New Roman','serif'; font-size: 16px;">Answered above

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