In a nuclear process, energy is released if binding energy per nucleon of the daughter products gets increased. In nuclear fission reaction, total binding energy of products formed due to nuclear fission is more than the parent fissionable material.
12.
The binding energy per nucleon of deuteron $$\left( {_1^2H} \right)$$ and helium nucleus $$\left( {_2^4He} \right)$$ is $$1.1\,MeV$$ and $$7\,MeV$$ respectively. If two deuteron nuclei react to form a single helium nucleus, then the energy released is
In sun, huge amount of energy is produced due to fusion of 4 protons (hydrogen nucleus) into a helium nucleus. According to the reaction
$$_1{H^1}{ + _1}{H^1}{ + _1}{H^1}{ + _1}{H^1}{ \to _2}H{e^4} + {2_{ + 1}}{\beta ^0} + \gamma \left( {{\text{energy}}} \right) + 2\upsilon $$
14.
The proton-proton mechanism that accounts for energy production in the sun releases $$26.7\,MeV$$ energy for each event. In this process, protons fuse to form an alpha particle $$\left( {^4He} \right).$$ At what rate $$\frac{{dm}}{{dt}}$$ hydrogen being consumed in the core of the sun by the $$p-p$$ cycle? Power of sun is $$3.90 \times {10^{26}}W.$$
The rate $$\frac{{dm}}{{dt}}$$ can be calculate as;
Power, $$P = \frac{{dE}}{{dt}} = \frac{{dE}}{{dm}} \times \frac{{dm}}{{dt}} = \frac{{\Delta E}}{{\Delta m}} \times \frac{{dm}}{{dt}}$$
$$\therefore \frac{{dm}}{{dt}} = \frac{{\Delta m}}{{\Delta E}}P\,.......\left( {\text{i}} \right)$$
We known that $$26.2\,MeV = 4.20 \times {10^{ - 12}}J$$ of thermal energy is produced when four protons are consumed. This is $$\Delta E = 4.20 \times {10^{ - 12}}J$$ for $$\Delta m = 4 \times \left( {1.67 \times {{10}^{ - 27}}kg} \right).$$
Substituting these values in equation (i), we have
$$\eqalign{
& \frac{{dm}}{{dt}} = \frac{{\Delta m}}{{\Delta E}}P = \frac{{4\left( {1.67 \times {{10}^{ - 27}}} \right)}}{{4.20 \times {{10}^{ - 12}}}} \times \left( {3.90 \times {{10}^{26}}} \right) \cr
& = 6.2 \times {10^{11}}kg/s \cr} $$
15.
In the nuclear fusion reaction
$$_1^2H + _1^3H \to _2^4He + n$$
given that the repulsive potential energy between the two
nuclei is $$ \sim 7.7 \times {10^{ - 14}}J,$$ the temperature at which the gases must be heated to initiate the reaction is nearly
[Boltzmann’s Constant $$k = 1.38 \times {10^{ - 23}}J/K$$ ]
The average kinetic energy per molecule $$ = \frac{3}{2}kT$$
This kinetic energy should be able to provide the repulsive potential energy
$$\eqalign{
& \therefore \frac{3}{2}kT = 7.7 \times {10^{ - 14}} \cr
& \Rightarrow T = \frac{{2 \times 7.7 \times {{10}^{ - 14}}}}{{3 \times 1.38 \times {{10}^{ - 23}}}} = 3.7 \times {10^9} \cr} $$
16.
If the nucleus $$_{13}^{27}Al$$ has a nuclear radius of about $$3.6\,fm,$$ then $$_{52}^{125}Te$$ would have its radius approximately as
If $$R$$ is the radius of the nucleus, the corresponding volume $$\frac{4}{3}\pi {R^3}$$ has been found to be proportional to $$A.$$
This relationship is expressed in inverse form as $$R = {R_0}{A^{\frac{1}{3}}}$$
The value of $${R_0}$$ is $$1.2 \times {10^{ - 15}}m,\,{\text{i}}{\text{.e}}{\text{.}}\,1.2\,fm$$
Therefore, $$\frac{{{R_{Al}}}}{{{R_{Te}}}} = \frac{{{R_0}{{\left( {{A_{Al}}} \right)}^{\frac{1}{3}}}}}{{{R_0}{{\left( {{A_{Te}}} \right)}^{\frac{1}{3}}}}}$$
$$\eqalign{
& \frac{{{R_{Al}}}}{{{R_{Te}}}} = \frac{{{{\left( {{A_{Al}}} \right)}^{\frac{1}{3}}}}}{{{{\left( {{A_{Te}}} \right)}^{\frac{1}{3}}}}} = \frac{{{{\left( {27} \right)}^{\frac{1}{3}}}}}{{{{\left( {125} \right)}^{\frac{1}{3}}}}} = \frac{3}{5} \cr
& {\text{or}}\,\,{R_{Te}} = \frac{5}{3} \times {R_{Al}} = \frac{5}{3} \times 3.6 = 6\,fm \cr} $$
17.
The mass number of $$He$$ is 4 and that for sulphur is 32. The radius of sulphur nuclei is larger than that of helium by
Volume of a nucleus is proportional to its mass number $$A.$$ If $$R$$ is the radius of the nucleus assumed to be spherical, then its volume $$\left( v \right) \propto A$$ (mass No.)
$$\eqalign{
& {\text{So,}}\,\,\left( {\frac{4}{3}\pi {R^3}} \right) \propto A \cr
& {\text{or}}\,\,R \propto {A^{\frac{1}{3}}} \cr
& \therefore \frac{{{R_S}}}{{{R_{He}}}} = {\left( {\frac{{{A_S}}}{{{A_{He}}}}} \right)^{\frac{1}{3}}} = {\left( {\frac{{32}}{4}} \right)^{\frac{1}{3}}} = 2 \cr
& {\text{or}}\,\,{R_S} = 2\,{R_{He}} \cr} $$
Fast neutrons can be easily slowed down by passing them through water.
19.
If the binding energy per nucleon in $$_3L{i^7}$$ and $$_2H{e^4}$$ nuclei are respectively $$5.60\,MeV$$ and $$7.06\,MeV,$$ then the energy of proton $$_3L{i^7} + p \to {2_2}H{e^4}$$ is
Total BE of nucleons in $$_3L{i^7} = 7 \times 5.60 = 39.20\,MeV$$
Total BE of nucleons in $$2\left( {_2H{e^4}} \right) = \left( {4 \times 7.06} \right) \times 2 = 56.48\,MeV$$
Therefore, energy of protons in the reaction = difference of BE's
$$ = 56.48 - 39.20 = 17.3\,MeV$$
20.
The constituents of atomic nuclei are believed to be
According to proton-neutron hypothesis, a nucleus of mass number $$A$$ and atomic number $$Z$$ contains $$Z$$ protons and $$\left( {A - Z} \right)$$ neutrons. Constituents of atomic nucleus are Nucleons i.e. neutron and proton.