Chapter 13: Nuclei

Ans: We know the expression for nuclear radius to be:

$R={{R}_{0}}{{A}^{\frac{1}{3}}}$

Where, ${{R}_{0}}$ is a Constant and $A$ is the mass number of the nucleus

Nuclear matter density would be,

$\rho =\frac{\text{Mass of the nucleus}}{\text{Volume of the nucleus}}$

Now, let m be the average mass of the nucleus, then, mass of the nucleus $=mA$

Nuclear density,

$\rho =\frac{mA}{\frac{4}{3}\pi {{R}^{3}}}=\frac{3mA}{4\pi {{\left( {{R}_{0}}{{A}^{\frac{1}{3}}} \right)}^{3}}}=\frac{3mA}{4\pi {{R}_{0}}^{3}A}$

$\therefore \rho =\frac{3m}{4\pi {{R}_{0}}^{3}}$

Therefore, we found the nuclear matter density to be independent of A and it is found to be nearly constant. 

Ans: Let the amount of energy released during the electron capture process be ${{Q}_{1}}$ . The nuclear reaction could be written as:

${{e}^{+}}+{}_{Z}^{A}X\to {}_{Z-1}^{A}Y+\nu +{{Q}_{1}}$………………………… (1)

Let the amount of energy released during the positron capture process be ${{Q}_{2}}$. The nuclear reaction could be written as:

${}_{Z}^{A}X\to {}_{Z-1}^{A}Y+{{e}^{+}}+\nu +{{Q}_{2}}$………………………… (2)

Let, ${{m}_{N}}\left( {}_{Z}^{A}X \right)\text{ be the nuclear mass of }{}_{Z}^{A}X$, ${{m}_{N}}\left( {}_{Z-1}^{A}Y \right)\text{ be the nuclear mass of }{}_{Z-1}^{A}Y$

$m\left( {}_{Z}^{A}X \right)\text{ be the atomic mass of }{}_{Z}^{A}X$

$m\left( {}_{Z-1}^{A}Y \right)\text{ be the nuclear mass of }{}_{Z-1}^{A}Y$

${{m}_{e}}\text{ be the mass of an electron}$, $c\text{ be the speed of light}$, then, the Q-value of the electron capture reaction could be given as, 

${{Q}_{1}}=\left[ {{m}_{N}}\left( {}_{Z}^{A}X \right)+{{m}_{e}}-{{m}_{N}}\left( {}_{Z-1}^{A}Y \right) \right]{{c}^{2}}$

$\Rightarrow {{Q}_{1}}=\left[ m\left( {}_{Z}^{A}X \right)-Z{{m}_{e}}+{{m}_{e}}-m\left( {}_{Z-1}^{A}Y \right)+\left( Z-1 \right){{m}_{e}} \right]{{c}^{2}}$

$\Rightarrow {{Q}_{1}}=\left[ m\left( {}_{Z}^{A}X \right)-m\left( {}_{Z-1}^{A}Y \right) \right]{{c}^{2}}$………………… (3)

The Q-value of the positron capture reaction could be given as, 

${{Q}_{2}}=\left[ {{m}_{N}}\left( {}_{Z}^{A}X \right)-{{m}_{N}}\left( {}_{Z-1}^{A}Y \right)-{{m}_{e}} \right]{{c}^{2}}$

$\Rightarrow {{Q}_{1}}=\left[ m\left( {}_{Z}^{A}X \right)-Z{{m}_{e}}-m\left( {}_{Z-1}^{A}Y \right)+\left( Z-1 \right){{m}_{e}}-{{m}_{e}} \right]{{c}^{2}}$

$\Rightarrow {{Q}_{1}}=\left[ m\left( {}_{Z}^{A}X \right)-m\left( {}_{Z-1}^{A}Y \right)-2{{m}_{e}} \right]{{c}^{2}}$………………… (4) 

It can be inferred that if \[{{Q}_{2}}>0\], then; Also, if \[Q1>0\], it does not necessarily mean that \[Q2>0\]. In other words, we could say that if ${{\beta }^{+}}$emission is energetically allowed, then the electron capture process is necessarily allowed, but not vice-versa. This is so because the Q-value must be positive for an energetically-allowed nuclear reaction.

Ans: We are given:

Average atomic mass of magnesium, \[m=24.312\text{ }u\]

Mass of magnesium ${}_{12}^{24}Mg$ isotope, \[{{m}_{1}}=23.98504\text{ }u\]

Mass of magnesium ${}_{12}^{25}Mg$ isotope, \[{{m}_{2}}=24.98584\text{ }u\]

Mass of magnesium ${}_{12}^{26}Mg$ isotope, \[{{m}_{3}}=25.98259\text{ }u\]

Abundance of ${}_{12}^{24}Mg,\text{ }{{\eta }_{1}}=78.99%$ 

Abundance of  ${}_{12}^{25}Mg,\text{ }{{\eta }_{2}}=x%$

Now, the abundance of ${}_{12}^{26}Mg$, \[{{\eta }_{3}}=100-x-78.99%=\left( 21.01-x \right)%\]

Also, we have the relation for the average atomic mass as:

$m=\frac{{{m}_{1}}{{\eta }_{1}}+{{m}_{2}}{{\eta }_{2}}+{{m}_{3}}{{\eta }_{3}}}{{{\eta }_{1}}+{{\eta }_{2}}+{{\eta }_{3}}}$

$\Rightarrow 24.312=\frac{23.98504\times 78.99+24.98584\times x+25.98259\times \left( 21.01-x \right)}{100}$

$\Rightarrow 0.99675x=9.2725255$

$\therefore x\approx 9.3%$

And, $21.01-x=11.71%$

Therefore, we found the abundance of ${}_{12}^{25}Mg$ to be 9.3% and that of ${}_{12}^{26}Mg$ to be 11.71%.

Ans: For a neutron removal from ${}_{20}C{{a}^{41}}$nucleus, the corresponding nuclear reaction could be written as, 

${}_{20}^{41}Ca\to {}_{20}^{40}Ca+{}_{0}^{1}n$

We are given:

$m\left( {}_{20}^{40}Ca \right)=39.962591u$

$m\left( {}_{20}^{41}Ca \right)\text{=40}\text{.962278u}$

$m\left( {}_{0}{{n}^{1}} \right)=1.008665u$

Now, the mass defect for this reaction could be given by, 

$\Delta m=m\left( {}_{20}^{40}Ca \right)+\left( {}_{0}^{1}n \right)-m\left( {}_{20}^{41}Ca \right)$

$\Rightarrow \Delta m=39.962591+1.008665-40.962278=0.008978u$

But we know, $1u=931.5MeV/{{c}^{2}}$

$\Rightarrow \Delta m=0.008978\times 931.5MeV/{{c}^{2}}$

Now, we could calculate the energy required for the neutron removal by, 

$E=\Delta m{{c}^{2}}$

$\Rightarrow E=0.008978\times 931.5=8.363007MeV$

For the case of ${}_{13}^{27}Al$, the neutron removal reaction could be written as, 

${}_{13}^{27}Al\to {}_{13}^{26}Al+{}_{0}^{1}n$

We are given,

$m\left( {}_{13}^{26}Al \right)=25.986995u$

$m\left( {}_{13}^{27}Al \right)=26.981541u$

Now, the mass defect here could be given by, 

$\Delta m=m\left( {}_{13}^{26}Al \right)+m\left( {}_{0}^{1}n \right)-m\left( {}_{13}^{27}Al \right)$

$\Rightarrow \Delta m=25.986895+1.008665-26.981541=0.014019u$

$\Rightarrow \Delta m=0.014019\times 931.5MeV/{{c}^{2}}$

Therefore, the energy that is required for the removal of neutron would be, 

$E=\Delta m{{c}^{2}}=0.014019\times 931.5$

$\therefore E=13.059MeV$

Ans: We are given:

Half life of  ${}_{15}^{32}P\text{ }\left( {{T}_{\frac{1}{2}}}=14.3d \right)$

Half life of  ${}_{15}^{33}P\text{ }\left( {{T}_{\frac{1}{2}}}=25.3d \right)$

Now, we know that nucleus decay is 10% of the total amount of decay.

Also, the source has initially 10% of ${}_{15}^{32}P$ nucleus and 90% of ${}_{15}^{32}P$ nucleus.

Suppose after t days, the source has 10% of ${}_{15}^{32}P$ nucleus and 90% of ${}_{15}^{33}P$ 

nucleus.

Initially we have:

Number of ${}_{15}^{33}P$ nucleus $=N$

Number of ${}_{15}^{32}P$ nucleus $=9N$

Finally:

Number of ${}_{15}^{33}P$ nucleus $=9N'$

Number of ${}_{15}^{32}P$ nucleus$=N'$

For ${}_{15}^{32}P$ nucleus, we could write the number ratio as:

$\frac{N'}{9N}={{\left( \frac{1}{2} \right)}^{\frac{t}{{{T}_{\frac{1}{2}}}}}}$

$\Rightarrow N'=9N{{\left( 2 \right)}^{\frac{-t}{14.3}}}$………………………. (1)

Now, for ${}_{15}^{33}P$, we could write the number ratio as, 

$\frac{9N'}{N}={{\left( \frac{1}{2} \right)}^{\frac{1}{T{{'}_{\frac{1}{2}}}}}}$

$\Rightarrow 9N'=N{{\left( 2 \right)}^{\frac{-t}{25.3}}}$…………………………. (2)

We could now divide equation (1) by equation (2) to get, 

$\frac{1}{9}=9\times {{2}^{\left( \frac{t}{25.3}-\frac{t}{14.3} \right)}}$

$\Rightarrow \frac{1}{81}={{2}^{\left( -\frac{11t}{25.3\times 14.3} \right)}}$

$\Rightarrow \log 1-\log 81=\frac{-11t}{25.3\times 14.3}\log 1$

$\Rightarrow \frac{-11t}{25.3\times 14.3}=\frac{0-1.908}{0.301}$

$\therefore t=\frac{25.3\times 14.3\times 1.908}{11\times 0.301}\approx 208.5days$

Therefore, we found that it would take about 208.5days for 90% decay of ${}_{15}^{33}P$.

Ans: Consider a ${}_{6}^{14}C$ emission nuclear reaction, 

${}_{88}^{223}Ra\to {}_{82}^{209}Pb+{}_{6}^{14}C$

We know that:

Mass of ${}_{88}^{223}Ra,\text{ }{{\text{m}}_{1}}=223.01850u$

Mass of ${}_{6}^{14}C,\text{ }{{\text{m}}_{3}}=14.00324u$

Now, the Q-value of the reaction could be given as:

$Q=\left( {{m}_{1}}-{{m}_{2}}-{{m}_{3}} \right){{c}^{2}}$

$\Rightarrow Q=\left( 223.01850-208.98107-14.00324 \right){{c}^{2}}=\left( 0.03419{{c}^{2}} \right)u$

But we have, $1u=931.5MeV/{{c}^{2}}$

$\Rightarrow Q=0.03419\times 931.5$

$\therefore Q=31.848MeV$

Hence, the Q-value of the nuclear reaction is found to be 31.848 MeV. Since the value is positive, the reaction is energetically allowed.

Now consider a ${}_{2}^{4}He$ emission nuclear reaction:

${}_{88}^{223}Ra\to {}_{86}^{229}Rn+{}_{2}^{4}He$

We know that:

Mass of ${}_{88}^{223}Ra,\text{ }{{\text{m}}_{1}}=223.01850$

Mass of ${}_{82}^{219}Rn,\text{ }{{\text{m}}_{2}}=219.00948$

Mass of ${}_{2}^{4}He,\text{ }{{\text{m}}_{3}}=4.00260$

Q-value of this nuclear reaction could be given as:

$Q=\left( {{m}_{1}}-{{m}_{2}}-{{m}_{3}} \right){{c}^{2}}$

$\Rightarrow Q=\left( 223.01850-219.00948-4.00260 \right){{c}^{2}}$

$\Rightarrow Q=\left( 0.00642{{c}^{2}} \right)u$

$\therefore Q=0.00642\times 931.5=5.98MeV$

Therefore, the Q-value of the second nuclear reaction is found to be 5.98MeV. Since the value is positive, we could say that the reaction is energetically allowed.

Ans: We are given:

In the fission of ${}_{92}^{238}U$, 10 $\beta -$particles decay from the parent nucleus. The nuclear reaction can be written as:

${}_{92}^{238}U+{}_{0}^{1}n\to {}_{58}^{140}Ce+{}_{44}^{99}Ru+10{}_{-1}^{0}e$

It is also given that:

Mass of a nucleus of ${}_{92}^{238}U,\text{ }{{\text{m}}_{1}}=238.05079u$

Mass of a nucleus of ${}_{58}^{140}Ce,\text{ }{{\text{m}}_{2}}=139.90543u$

Mass of nucleus of ${}_{44}^{99}Ru,\text{ }{{\text{m}}_{3}}=98.90594u$, 

Mass of a neutron ${}_{0}^{1}n,\text{ }{{\text{m}}_{4}}=1.008665u$

Q-value of the above equation would be,

$Q=\left[ m'\left( {}_{92}^{238}U \right)+m\left( {}_{0}^{1}n \right)-m'\left( {}_{58}^{140}Ce \right)-m'\left( {}_{44}^{99}Ru \right)-10{{m}_{e}} \right]{{c}^{2}}$

Where, $m'=$Represents the corresponding atomic masses of the nuclei

$m'\left( {}_{92}^{238}U \right)={{m}_{1}}-92{{m}_{e}}$

$m'\left( {}_{58}^{140}Ce \right)={{m}_{2}}-58{{m}_{e}}$

$m'\left( {}_{44}^{99}Ru \right)={{m}_{3}}-44{{m}_{e}}$

$m\left( {}_{0}^{1}n \right)={{m}_{4}}$

$Q=\left[ {{m}_{1}}-92{{m}_{e}}+{{m}_{4}}-{{m}_{2}}+58{{m}_{e}}-{{m}_{3}}+44{{m}_{e}}-10{{m}_{e}} \right]{{c}^{2}}$

$\Rightarrow Q=\left[ {{m}_{1}}+{{m}_{4}}-{{m}_{2}}-{{m}_{3}} \right]{{c}^{2}}=\left[ 238.0507+1.008665-139.90543-98.90594 \right]{{c}^{2}}$

$\Rightarrow Q=\left[ 0.247995{{c}^{2}} \right]u$

But $1u=931.5MeV/{{c}^{2}}$

\[\therefore Q=0.247995\times 931.5\text{ =}231.007MeV\]

Therefore, the Q-value of the fission process is found to be 231.007MeV.

Ans: Consider the D-T nuclear reaction, 

${}_{1}^{2}H+{}_{1}^{3}H\to {}_{2}^{4}He+n$

We are also given that:

Mass of \[{}_{1}^{2}H,\text{ }{{\text{m}}_{1}}=2.014102u\]

Mass of \[{}_{1}^{3}H,\text{ }{{\text{m}}_{2}}=3.016049u\]

Mass of ${}_{2}^{4}He,\text{ }{{\text{m}}_{3}}=4.002603u$

Mass of ${}_{0}^{1}n,\text{ }{{\text{m}}_{4}}=1.008665u$

Now, the Q-value of the given D-T reaction would be:

$Q=\left[ {{m}_{1}}+{{m}_{2}}-{{m}_{3}}-{{m}_{4}} \right]{{c}^{2}}$

$\Rightarrow Q=\left[ 2.014102+3.016049-4.002603-1.008665 \right]{{c}^{2}}$

$\Rightarrow Q=\left[ 0.018883{{c}^{2}} \right]u$

But $1u=931.5MeV/{{c}^{2}}$

$\therefore Q=0.018883\times 931.5=17.59MeV$

1. Consider the radius of both deuterium and tritium to be approximately 2.0fm. What is the kinetic energy needed to overcome the coulomb repulsion between the two nuclei? To what temperature must the gas be heated to initiate the reaction? (Hint: Kinetic energy required for one fusion event = average thermal kinetic energy available with the interacting particles $2\left( \frac{3kT}{2} \right)$; k = Boltzmann’s constant, T = absolute temperature.)

Ans: We are given:

Radius of deuterium and tritium, $r\approx 2.0fm=2\times {{10}^{-15}}m$ 

Distance between the two nuclei at the moment when they touch each other,

$d=r+r=4\times {{10}^{-15}}m$

Charge on the deuterium nucleus $=e$

Charge on the tritium nucleus $=e$

Hence, the repulsive potential energy between the two nuclei could be given as:

$V=\frac{{{e}^{2}}}{4\pi {{\varepsilon }_{0}}d}$

Where, ${{\varepsilon }_{0}}=$Permittivity of free space

$\frac{1}{4\pi {{\varepsilon }_{0}}}=9\times {{10}^{9}}N{{m}^{2}}{{c}^{-2}}$

$\Rightarrow V=\frac{9\times {{10}^{9}}\times {{\left( 1.6\times {{10}^{-19}} \right)}^{2}}}{4\times {{10}^{-15}}}=5.76\times {{10}^{-14}}J=\frac{5.76\times {{10}^{-14}}}{1,6\times {{10}^{-19}}}$

$\therefore V=3.6\times {{10}^{5}}eV=360keV$

Therefore, $5.76\times {{10}^{-14}}J$ or  $360keV$of kinetic energy (KE) is needed to overcome the coulomb repulsion between the two nuclei.

However, we are also given that:

$KE=2\times \frac{3kT}{2}$

Where, $k=$Boltzmann constant 

$T=$Temperature required for triggering the reaction

$\therefore T=\frac{KE}{3k}=\frac{5.76\times {{10}^{-14}}}{3\times 1.38\times {{10}^{-23}}}=1.39\times {{10}^{9}}K$

Therefore, we found that the gas must be heated to a temperature of $1.39\times {{10}^{9}}K$ to initiate the reaction.

Ans: It can be observed from the given $\gamma $-decay diagram that ${{\gamma }_{1}}$decays from the 1.088MeV energy level to the 0MeV energy level. Hence, the energy corresponding to ${{\gamma }_{1}}$-decay is given as:

${{E}_{1}}=1.088-8=1.088MeV$

$\Rightarrow h{{\nu }_{1}}=1.088\times 1.6\times {{10}^{-19}}\times {{10}^{6}}J$ 

Where, Planck's constant  $h=6.6\times {{10}^{-34}}Js$

${{\nu }_{1}}=$Frequency of radiation radiated by ${{\gamma }_{1}}-$decay

${{\nu }_{1}}=\frac{{{E}_{1}}}{h}$

$\Rightarrow {{\nu }_{1}}=\frac{1.088\times 1.6\times {{10}^{-19}}\times {{10}^{6}}}{6.6\times {{10}^{-34}}}=2.637\times {{10}^{20}}Hz$

It can be observed from the given $\gamma -$decay diagram that ${{\gamma }_{2}}$decays from the \[0.412MeV\]energy level to the \[0MeV\]energy level.

Now, the energy corresponding to ${{\gamma }_{2}}$-decay could be given as:

${{E}_{2}}=0.412-0=0.412MeV$

$\Rightarrow h{{\nu }_{2}}=0.412\times 1.6\times {{10}^{-19}}\times {{10}^{6}}J$

Where, ${{\nu }_{2}}=$Frequency of radiation radiated by ${{\gamma }_{2}}-$decay

${{\nu }_{2}}=\frac{{{E}_{2}}}{h}=\frac{0.412\times 1.6\times {{10}^{-19}}\times {{10}^{6}}}{6.6\times {{10}^{-34}}}=9.988\times {{10}^{19}}Hz$

It can be observed from the given $\gamma $-decay diagram that ${{\gamma }_{3}}$-decays from the \[1.088MeV\]energy level to the \[0.412MeV\]energy level.

Now, the energy corresponding to ${{\gamma }_{3}}$ -decay is given as:

${{E}_{3}}=1.088-0.412=0.676MeV$

$\Rightarrow h{{\nu }_{3}}=0.676\times {{10}^{-19}}\times {{10}^{6}}$

Where, ${{\nu }_{3}}=$Frequency of radiation radiated by ${{\gamma }_{3}}-$decay

${{\nu }_{3}}=\frac{{{E}_{3}}}{h}=\frac{0.676\times 1.6\times {{10}^{-19}}\times {{10}^{6}}}{6.6\times {{10}^{-34}}}=1.639\times {{10}^{20}}Hz$

Mass of \[m\left( {}_{78}^{198}Au \right)\text{ =}197.968233u\]

Mass of $m\left( {}_{80}^{198}Hg \right)=197.966760u$

\[1u=931.5MeV/{{c}^{2}}\]

Energy of the highest level could be given as:

$E=\left[ m\left( {}_{78}^{198}Au \right)-m\left( {}_{80}^{190}Hg \right) \right]=197.968233-197.966760=0.001473u$

$\Rightarrow E=0.001473\times 931.5=1.3720995MeV$

${{\beta }_{1}}$decays from the \[1.3720995MeV\]level to the \[1.088MeV\]level

Maximum kinetic energy of the ${{\beta }_{1}}$particle $=1.3720995\text{ }-\text{ }1.088$

$\Rightarrow K.E=0.2840995MeV$

${{\beta }_{2}}$ decays from the $1.3720995MeV$level to that of the $0.412MeV$level. Now, we find the maximum kinetic energy of the ${{\beta }_{2}}$particle to be, 

$K.{{E}_{\max }}=1.3720995-0.412=0.9600995MeV$

Therefore, we found the maximum kinetic energy of the ${{\beta }_{2}}$particle to be $0.9600995MeV$. 

Ans: We are given:

Amount of hydrogen, \[m=1\text{ }kg=1000g\]

1 mole, i.e., 1g of hydrogen $\left( {}_{1}^{1}H \right)$ contains \[6.023\times {{10}^{23}}atoms\].

That is, 1000g of ${}_{1}^{1}H$ contains \[6.023\times {{10}^{23}}atoms\].

Within the sun, four ${}_{1}^{1}H$ nuclei combine and form one ${}_{2}^{4}He$ nucleus. In this process \[26MeV\]of energy is released.

Hence, the energy released from the fusion of 1 kg ${}_{1}^{1}H$ is:

${{E}_{1}}=\frac{6.023\times {{10}^{23}}\times 26\times {{10}^{3}}}{4}=39.1495\times {{10}^{26}}MeV$

Therefore, we found the energy released during the fusion of 1kg ${}_{1}^{1}H$ is:

${{E}_{1}}=\frac{6.023\times {{10}^{23}}\times 26\times {{10}^{3}}}{4}=39.1495\times {{10}^{26}}MeV$

Hence, the energy released during the fusion of 1kg of ${}_{1}^{1}H$ to be $39.1495\times {{10}^{26}}MeV$. 

1. The Fission of 1.0kg of ${}^{235}U$ in a fission reactor.

Ans: We are given:

Amount of ${}_{92}{{U}^{235}}=1000gm$

1 mole, i.e., 235g of ${}_{92}^{235}U$ contains \[6.023\times {{10}^{23}}atoms\].

1000g of ${}_{92}^{235}U$ contains $\frac{6.023\times {{10}^{23}}\times 1000}{235}atoms$

We know that the amount of energy released in the fission of one atom of ${}_{92}^{235}U$ is\[200MeV\]. Therefore, energy released from the fission of 1kg of ${}_{92}^{235}U$ is:

${{E}_{2}}=\frac{6\times {{10}^{23}}\times 1000\times 200}{235}=5.106\times {{10}^{26}}MeV$

$\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{39.1495\times {{10}^{26}}}{5.106\times {{10}^{26}}}=7.67\approx 8$

Hence, we found the energy released during the fusion of 1kg of hydrogen is nearly 8 times the energy released during the fusion of 1kg of uranium.