How long can an electric lamp of 100 W be kept glowing by fusion of 2.0 kg of deuterium? Take the fusion reaction as :\$^2_1H\, +\, ^2_1\, \rightarrow\, ^3_2He\, +\, n\, +\, 3.27\, MeV\$.

Number of atoms in 2 kg deuterium = \$\dfrac{6.023\times 10^{23}\times 2000}{2}=6.023\times 10^{26}\$Energy released when 2 atoms fuse = \$3.27MeV\$Thus, total energy released = \$\dfrac{3.27}{2}\times 6.023\times 10^{26}MeV\$\$=15.75\times 10^{13}J\$.Energy consumed by bulb per second=\$100J\$.Thus, time for which the bulb glows = \$\dfrac{15.75\times 10^{13}}{100}s\$\$=15.75\times 10^{11}s=4.99\times 10^7 years\$.

Obtain the binding energy of a nitrogen nucleus \$(_7N^{14})\$ from the following data: \$m_H=1.00783\$ a.m.u., \$m_n=1.0087\$ a.m.u.; \$m_N=14.00307\$ a.m.u.

\$m_{H}\$ = 1.00783 amu\$m_{n}\$ = 1.0087 amu\$m_{N}\$ = 14.0030u amu In\$_{7}N^{14}\$, these are 7 protons and 7 neutrons. \$\therefore \$ mass defect,\$ \Delta =( 7m_{H}+7m_{n}) - m_{N} \$=(7×1.00783)+(7×1.00807)-14.00307=0.11243 amuBinding energy of nitrogen nucleus= \$\Delta\$×931 MeV =0.11243×931=104.67

The binding energy per nucleon of \$^7_3 Li\$ and \$^4_2 He\$ nuclei are 5.60 MeV and 7.06 Me V, respectively. In the nuclear reaction \$^7_3 Li + ^1_1 He \longrightarrow ^4_2 H + ^4_2 He + Q\$ the value of energy Q released is

The binding energy for \$_1H^1\$ is around zero and also not give in the question so we can ignore it,\$Q=2(4\times 7.06)-7\times (5.60)\$\$=(56.48-39.2)MeV\$\$=17.28\ MeV\$\$=17.3\ MeV\$

Calculate the binding energy and binding energy per nucleon (in MeV) of a nitrogen nucleus \$(^{14}_7N)\$ from the following data :Mass of proton = 1.00783 uMass of neutron = 1.00867 uMass of nitrogen nucleus = 14.00307 u

Nuclear reaction : \$7p+7n\rightarrow ^{14}_7 N\$Mass defect \$\Delta M = 7m_p + 7m_n - m_{nucleus}\$\$\therefore\$ \$\Delta M = 7(1.00783)+7(1.00867) - 14.00307\$ amu\$\implies\$ \$\Delta M = 0.11243\$ amuThus binding energy \$E = \Delta M \times 931.5\$ MeV\$\therefore\$ \$E = 0.11243\times 931.5 = 104.73\$ MeVBinding energy per nucleon \$= \dfrac{104.73}{14} = 7.48\$ MeV

Draw the graph showing the variation of binding energy per nucleon with mass numbers. Give the reason for the decrease of binding energy per nucleon for nuclei with higher mass number.

The graph of the binding energy per nucleon versus mass number A is shown in figure. The decrease of the binding energy per nucleon for nuclei with high mass number is due to increased coulomb repulsion between protons inside the nucleus.

Consider the following statements (A) and (B) and identify the correct answer given below :Statement A: Positive values of packing fraction implies a large value of binding energyStatement B: The difference between the mass of the nucleus and the mass number of the nucleus is called the packing fraction

Packing fraction \$= \dfrac{Actual \ isotopic \ mass - mass \ number}{ Mass\ number}\$ A negative packing fraction indicates stability of the nucleus thus a large Binding energy and vice versa. So, we easily see that statement A and b are false.

Obtain the binding energy of the nuclei \$_{26}Fe^{56}\$ and \$_{83}Bi^{209}\$ in units of MeV from the following data: mass of hydrogen atom \$=1.007825\$ a.m.u., mass of neutron \$=1.08665\$ a.m.u.; mass of \$_{26}Fe^{56}\$ atom \$=55.934939\$ a.m.u. and mass of \$_{83}Bi^{209}\$ atom \$=208.980388\$a.m.u. Also calculate binding energy per nucleon in the two cases.

(i) \$_{26}Fe^{56}\$ nucleus contains 26 protons .Number of neutrons \$=(56-26) = 30 \$ neutronsNow,Mass of 26 protons\$ = 26\$ X \$1.007825 = 26.20345 u\$Mass of 30 neutrons\$ = 30\$ X \$1.008665= 30.25995 u\$Total mass of 56 nucleons\$= 56.46340 u\$Mass of \$_{26}Fe{56}\$ nucleus\$= 55.934939 u\$Therefore,Mass defect, \$\triangle m\$ \$= 56.46340-55.934939 = 0.528461 u\$Total Binding Energy\$= 0.528461\$ X \$931.5 Mev = 492.26 MeV\$Average binding energy per nucleon\$= \frac{492.26}{56}=8.790 MeV\$ (ii) \$_{83}Bi^{209}\$ nucleus contains 83 protonsNumber of neutrons\$=209-83=126 neutrons\$Now,Mass of 83 protons\$= 83\$ X \$1.007825= 83.649475 amu\$Mass of 126 neutrons\$ = 126 X 1.008665= 127.091790 amu\$Therefore, total mass of nucleons\$= 83.649475+127.091790=210.741260amu\$Given, mass of nucleus\$=208.980388 amu\$Now, mass defect \$\triangle m\$ \$=210.741260-208.980388=1.760872\$Total binding energy\$=1.760872\$ X \$931.5 = 1640.26 MeV\$Therefore, average binding energy per nucleon = \$\frac{1640.26}{209}=7.848 Mev\$

(a) Draw the plot of blinding energy per nucleon (BE/A)as a function of mass number. Write two important conclusion that can be drawn regarding the nature of nuclear source.(b) Use this graph to explain the release of energy in both the process of nuclear fusion and fission.(c) Write the basic nuclear process of neutron undergoing \$\beta \$-decay. why is the detection of neutrinos found very difficult?

(a)Conclusions drawn from the graph:(i) The nuclear force is very attractive and strong to produce binding energy of the order of MeV per nucleon.(ii)The nuclear force is a short range force which can be shown by its value being constant in the range 30<A<170.(b)In nuclear fission a heavier nucleus breaks into lighter nuclei to release high binding energy per nucleon. In nuclear fusion, the binding energy of the fused heavier nucleon is more than the binding energy of the lighter nuclei which combined to form the heavier nuclei. Energy is also released in nuclear fusion.(c)the basic nuclear process of neutron undergoing \$\beta\$ decay is:\$n\rightarrow p+\vec e+\vec \nu\$Neutrinos have high penetrating power which makes their detection difficult.

The mass of \$_{3}Li^{7}\$ nucleus is \$0.042\$ a.m.u. less than the sum of masses of its nucleons. Find the binding energy per nucleon.

We have Mass number = 7 and Atomic number= 3If mass, \$m = 1 u\$ and coulomb force \$c= 3\$X\$10^8 ms^{-1}\$, then\$E=931 MeV\$ that is, \$1u=931 MeV\$Binding Energy = \$0.042\$X\$931=39.10 MeV\$Therefore, BInding Energy per Nucleon \$= \dfrac{ Binding Energy}{Mass Number}=\dfrac{39.10}{7}= 5.58= 5.6MeV\$

Binding energy per nucleon vs. mass number curve for nuclei is shown in fig. W, X, Y and Z are four nuclei indicated on the curve. The process that would release energy is

Binding energy of \$W\$, \$|E_W| = 120\times7.5=900\$ MeVBinding energy of \$Y\$, \$|E_Y| = 30\times8.5=255\$ MeVIn process : \$W\rightarrow 2Y\$\$Q\$ value, \$Q = |E_W| - |E_Y| = 900-2(255) =390\$ MeVAs \$Q\$ value is positive, thus energy is released in process \$W\rightarrow 2Y\$

Nuclear Binding Energy | Definition, Examples, Diagrams

Q: The mass of \$_{3}Li^{7}\$ nucleus is \$0.042\$ a.m.u. less than the sum of masses of its nucleons. Find the binding energy per nucleon.

We have Mass number = 7 and Atomic number= 3If mass, \$m = 1 u\$ and coulomb force \$c= 3\$X\$10^8 ms^{-1}\$, then\$E=931 MeV\$ that is, \$1u=931 MeV\$Binding Energy = \$0.042\$X\$931=39.10 MeV\$Therefore, BInding Energy per Nucleon \$= \dfrac{ Binding Energy}{Mass Number}=\dfrac{39.10}{7}= 5.58= 5.6MeV\$

Q: Binding energy per nucleon vs. mass number curve for nuclei is shown in fig. W, X, Y and Z are four nuclei indicated on the curve. The process that would release energy is

Binding energy of \$W\$, \$|E_W| = 120\times7.5=900\$ MeVBinding energy of \$Y\$, \$|E_Y| = 30\times8.5=255\$ MeVIn process : \$W\rightarrow 2Y\$\$Q\$ value, \$Q = |E_W| - |E_Y| = 900-2(255) =390\$ MeVAs \$Q\$ value is positive, thus energy is released in process \$W\rightarrow 2Y\$

definition

Mass defect

The difference in mass of a nucleus and its constituents,

Δ M

, is called mass defect and is given by

Δ M = [Z m_{p} + (A - Z) m_{n}] - M

definition

Variation in binding energy per nucleon with mass number

The force is attractive and sufficiently strong to produce a binding energy of a few MeV per nucleon.
The constancy of the binding energy in the range 30 < A < 170 is a consequence of the fact that the nuclear force is short-ranged.
A very heavy nucleus, say A = 240, has lower binding energy per nucleon compared to that of a nucleus with A = 120. Thus if a nucleus A = 240 breaks into two A = 120 nuclei, nucleons get more tightly bound. This implies energy would be released in the process.
Consider two very light nuclei (A $\leq$ 10) joining to form a heavier nucleus. The binding energy per nucleon of the fused heavier nuclei is more than the binding energy per nucleon of the lighter nuclei.This means that the final system is more tightly bound than the initial system. Again energy would be released in such a process of fusion.

definition

Binding energy and binding energy per nucleon

If a certain number of neutrons and protons are brought together to form a nucleus of a certain charge and mass, an energy

E_{b}

will be released in the process. The energy

E_{b}

is called the binding energy of the nucleus.

E_{b} = Δ M c^{2}

The ratio of the binding energy

E_{b}

of a nucleus to the number of the nucleons, A, in that nucleus is called the binding energy per nucleon,

E_{b n}

E_{b n} = \frac{E _{b}}{A}

definition

Exoergic and Endoergic Reactions

Nuclear reactions are reactions in which either two nuclei collide with one another or a small particle such as a neutron collides with the nucleus of some atom.
Nuclear reactions can be described using energy changes as follow :
An exothermic (exoergic) nuclear reaction is a reaction that releases energy while an endothermic (or endoergic) reaction is one that requires an input of energy to take place.
The changes in kinetic energy during a nuclear reaction is defined as the reaction energy (or Q-value).
If the reaction energy is positive, the reaction is exothermic but if the reaction energy is negative, the reaction is endothermic.

Nuclear Binding Energy

definition

Mass defect

definition

Variation in binding energy per nucleon with mass number

definition

Binding energy and binding energy per nucleon

definition

Exoergic and Endoergic Reactions

Quick Summary With Stories