Calculate the binding energy per nucleon of 26Fe56. Atomic mass of 26Fe56 is 55.9349 u and that of 1H1 is 1.00783u. Mass of 0n1 = 1.00867u and 1u = 931 MeV. Solution: Given, Atomic mass of 26Fe56 (M) = 55.9349 u. Here, Z = 26 & A = 56 Mass of 1H1 (mp) = 1.00783 u. Mass of 0n1 (mn) = 1.00867 u 1u = 931 MeV We know that, B.E. = [Zmp + (A – Z) mn – M]. 931 MeV = [latex] [26\times 1.00783 + (56 – 26)\times 1.00867 – 55.9349]\times 931 MeV [/latex] = 492.294 MeV Now, B.E. per nucleon [latex] (\bar{B.E.}) = \frac{B.E.}{A} = \frac{492.294}{56} [/latex] = 8.791 Mev/nucleon.
Calculate the binding energy per nucleon of calcium nucleus (20Ca40). Given: mass of 20Ca40 = 39.962589 u Mass of neutron, mn = 1.008665u Mass of proton, mp = 1.007825u 1u = 931 MeV. Solution: Given, Atomic mass of 20Ca40 (M) = 39.962589 u. Here, Z = 20 & A = 40 Mass of 1H1 (mp) = 1.007825 u. Mass of 0n1 (mn) = 1.008665 u 1u = 931 MeV We know that, B.E. = [Zmp + (A – Z) mn – M]. 931 MeV = [latex][20\times 1.007825 + (40 – 20)\times 1.008665 – 39.962589]\times 931 MeV [/latex] = 341.873 MeV Now, B.E. per nucleon [latex](\bar{B.E.}) = \frac{B.E.}{A} = \frac{341.873}{40}[/latex] = 8.547 Mev/nucleon.
A city requires 107 watts of electrical power on the average. If this is to be supplied by a nuclear reactor of efficiency 20%. Using 92U235 as the fuel source, calculate the amount of fuel required per day. (Energy released per fission 92U235 = 200 MeV). Solution: Given, Electrical power required by city (Pout) = 107 Watt Efficiency [latex](\eta)[/latex] = 20% Amount of fuel required per day (m) = ? Energy released per fission (E) = 200 MeV = 200 x 1.6 x 10-13 = 3.2 x 10-11 J We have, efficiency [latex] (\eta) = \frac{P_{out}}{P_{in}}\times 100% [/latex]
Now, Time (t) = 1 day = 86400 Sec So, Power = [latex]\frac{Total\ Energy\ (E_T)}{Time\ taken\ (t)}[/latex]
Or, ET = Pin . t = 5 x 107 x 86400 = 4.320 x 1012 J Now, No. of atom released per fission (N) = [latex]\frac{E_T}{E} = \frac{4.320\times 10^{12}}{3.2\times 10^{-11}}[/latex] = [latex]1.350\times 10^{23}[/latex] Using Avogadro’s Hypothesis, ∵ 6.023 x 1023 atom of U – 235 is contained in 235 g of it. ∴ 1 atom of U – 235 is contained in [latex]\frac{235}{6.023\times 10^{23}}[/latex] g of it. ∴ 1.35 x 1023 atom of U – 235 is contained in [latex]\frac{235}{6.023\times 10^{23}}\times 1.35\times 10^{23}[/latex] g of it. = 0.0527 Kg of it.
A nucleus of 92U238 disintegrates according to 92U238→90Th234 + 2He4 Calculate: (i) the total energy released in the disintegration process. (ii) the K.E. of the [latex]\alpha[/latex] – particles, the nucleus at rest before disintegration. (Mass of 92U238 = [latex]3.859\times 10^{-25}[/latex] kg, mass of 90Th234 = [latex]3.787\times 10^{-25}[/latex] kg, mass of 2He4 = [latex]6.648\times 10^{-27}[/latex] kg). Solution: Given, 92U235 disintegrates according to 92U238 → 90Th234 + 2He4 We have, Mass of 92U238 = 3.859 x 10-25 Kg Mass of 90Th234 = 3.787 x 10-25 Kg Mass of 2He4 = 6.648 x 10-27 Kg Total energy released in the disintegration process (E) = [latex][\sum_{reactant} – \sum_{product}]\times c^2[/latex] = [3.859 x 10-25 – (3.787 x 10-25 + 6.648 x 10-27)] x c2 = 4.968 x 10-11 J = 310.5 MeV Now, K.E. of [latex]\alpha[/latex] – particles = [latex]\frac{M_{Th}}{M_{Th} + M_{He}}\times E[/latex]
The mass of 17Cl35 is 34.9800 amu. Calculate its binding energy and binding energy per nucleon. Mass of one proton = 1.007825 amu and mass of one neutron = 1.008665 amu. Solution: Given, Atomic mass of 17Cl35 (M) = 34.9800 u Here, Z = 17 & A = 35 Mass of proton (mp) = 1.007825 u. Mass of neutron (mn) = 1.008665 u 1u = 931 MeV We know that, B.E. = [Zmp + (A – Z) mn – M]. 931 MeV = [latex][17\times 1.007825 + (35 – 17)\times 1.008665 – 34.98]\times 931 MeV [/latex] = 287.674 MeV Now, Binding Energy per nucleon [latex](\bar{B.E.}) = \frac{B.E.}{A} = \frac{287.674}{35}[/latex] = 8.219 Mev/nucleon.
The energy liberated in the fission of single uranium – 235 atom is [latex]3.2\times 10^{-11} J[/latex]. Calculate the power production corresponding to the fission of 1 gm of uranium per day. Assume Avogadro constant as [latex]6\times 10^{23} mol^{-1}[/latex]. Solution: Given, Energy liberated in the fission of single uranium atom (E) = 3.2 x 10-11 J Power production (P) =? Mass of uranium (m) = 1 gm = 1 x 10-3 kg Time (t) = 1 day = 86400 s We know that, According to Avogadro’s hypothesis, [latex]\because[/latex] 235 gm of U – 235 contains [latex]6.023\times 10^{23}[/latex] atoms. [latex]\therefore[/latex] 1 gm of U – 235 contains [latex]\frac{6.023\times 10^{23}\ atoms}{235}[/latex]. Or, N = 2.563 x 1021 atoms. Total energy liberated (T.E.) = N.E = (2.563 x 1021) x (3.2 x 10-11) = 8.202 x 1010 J Power production (P) = [latex]\frac{T.E.}{t}[/latex] = [latex]\frac{8.202\times 10^{10}}{86400}[/latex] = 9.493 x 105 Watt.
What will be the amount of energy released in the fusion of three alpha particles into a C12 nucleus if mass of He4 and C12 nuclei are respectively 4.00263 amu and 12 amu. Solution: Given, Amount of energy released (E) =? Mass of He4 = 4.00263 u Mass of C12 = 12 u We have, the reaction involved in the fusion is written as: 32He4 → 6C12 + Q So, Q – energy = [latex][\sum_{reactant} – \sum_{product}]\times 931 MeV[/latex] = [(3 x 4.00263) – 12] x 931 MeV = 7.346 MeV.
The mass of the nucleus of the isotope Lithium 3Li7 is 7.014351 u. Find its binding energy and binding energy per nucleon. (Given mass of proton = 1.00727 u, mass of neutron = 1.008665 u). Solution: Given, Atomic mass of 3Li7 (M) = 7.014351 u Here, Z = 3 & A = 7 Mass of proton (mp) = 1.00727 u Mass of neutron (mn) = 1.008665 u 1u = 931 MeV We know that, B.E. = [Zmp + (A – Z) mn – M]. 931 MeV = [latex][3\times 1.00727 + (7 – 3)\times 1.008665 – 7.014351]\times 931 MeV[/latex] = 39.213 MeV Now, Binding Energy per nucleon [latex](\bar{B.E.}) = \frac{B.E.}{A} = \frac{39.213}{7}[/latex] = 5.602 Mev/nucleon.
28Ni62 may be described as the most strongly bound nucleus because it has the highest B.E. per nucleon. Its neutral atomic mass is 61.928349 amu. Find its mass defect, its total binding energy and binding energy per nucleon. Given, mass of neutron = 1.008665 amu Mass of proton = 1.007825 amu 1 amu = 931.5 MeV. Solution: Given, Atomic mass of 28Ni62 (M) = 61.928349 u Here, Z = 28 & A = 62 Mass of proton (mp) = 1.007825 u. Mass of neutron (mn) = 1.008665 u 1u = 931.5 MeV We know that, Mass defect [latex](\Delta m)[/latex] = [Zmp + (A – Z) mn – M] = [latex][28\times 1.007825 + (62 – 28)\times 1.008665 – 61.928349][/latex] = 0.585361 amu B.E. = [latex][\Delta m][/latex]. 931 MeV = [latex][0.585361]\times 931.5\ MeV[/latex] = 545.264 MeV Now, Binding Energy per nucleon [latex](\bar{B.E.}) = \frac{B.E.}{A} = \frac{545.264}{62}[/latex] = 8.795 Mev/nucleon.
