Chapter 13: Nuclei

Ans: We know that, 

Nuclear radius of the gold isotope ${}_{79}A{{u}^{197}}={{R}_{Au}}$

Nuclear radius of the silver isotope ${}_{47}A{{g}^{107}}={{R}_{Ag}}$

Mass number of gold, ${{A}_{Au}}=197$

Mass number of silver, ${{A}_{Ag}}=107$

We also know that the ratio of the radii of the two nuclei is related with their mass numbers as:

$\frac{{{R}_{Au}}}{{{R}_{Ag}}}={{\left(\frac{{{A}_{Au}}}{{{A}_{Ag}}} \right)}^{\frac{1}{3}}}=1.2256$

Hence, the ratio of the nuclear radii of the gold and silver isotopes is found to be about 1.23.

Ans: We know that, 

Alpha particle decay of ${}_{88}^{26}Ra$ emits a helium nucleus. As a result, its mass number reduces to $222=\left( 226-4 \right)$and its atomic number reduces to$86=\left( 88-2 \right)$. This is shown in the following nuclear reaction:

${}_{88}^{226}Ra\to {}_{86}^{222}Ra+{}_{2}^{4}He$

\[Q-value\text{ }of\text{ }emitted\text{ }\alpha -particle\text{ }=\text{ }\left( Sum\text{ }of\text{ }initial\text{ }mass\text{ }-\text{ }Sum\text{ }of\text{ }final\text{ }mass \right){{c}^{2}}\]Where, c = Speed of light 

It is also given that:

$m\left( {}_{88}^{226}Ra \right)=226.02540u$

$\text{m}\left( {}_{86}^{220}Rn \right)=220.01137u$

$m\left( {}_{2}^{4}He \right)=4.002603u$

On substituting these values into the above equation, 

$\text{Q value =}\left[ 226.02540-\left( 222.01750+4.002603 \right) \right]u{{c}^{2}}$

$Q\text{ value=0}\text{.005297u}{{\text{c}}^{2}}$

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

$\Rightarrow Q=0.005297\times 931.5\approx 4.94MeV$

Kinetic energy of the $\alpha $particle$=\left( \frac{\text{Mass number after decay}}{Mass\text{ number before decay}} \right)\times Q$

$\therefore K.{{E}_{\alpha }}=\frac{222}{226}\times 4.94=4.85MeV$

Hence, the Kinetic energy of the alpha particle is found to be $4.85MeV$.

1. ${}_{86}^{220}Rn$

Ans: We know that, Alpha particle decay of ${}_{86}^{220}Rn$ could be given as, 

${}_{86}^{220}Rn\to {}_{84}^{216}Po+{}_{2}^{4}He$

We are also given, 

Mass of ${}_{86}^{220}Rn=220.01137u$

Mass of ${}_{84}^{216}Po=216.00189u$

Now, Q value could be given as, 

$Q-value=\left[ 220.01137-\left( 216.00189+4.00260 \right) \right]\times 931.5\approx 641MeV$

Now, we have the kinetic energy as, 

$K.{{E}_{\alpha }}=\left( \frac{220-4}{220} \right)\times 6.41=6.29MeV$

The kinetic energy of the alpha particle is found to be $6.29MeV$. 

Ans: The given nuclear reaction is, 

${}_{6}^{11}C\to {}_{5}^{11}B+{{e}^{+}}+\nu $

Half life of ${}_{6}^{11}C$ nuclei, ${{T}_{\frac{1}{2}}}=20.3\min $

Atomic masses are given to be:

$m\left( {}_{6}^{11}C \right)=11.011434u$

$m\left( {}_{6}^{11}B \right)=11.009305u$

Maximum energy that is possessed by the emitted positron would be \[0.960MeV\]. The change in the Q - value \[\left( \Delta Q \right)\] of the nuclear masses of the ${}_{6}^{11}C$

$\Delta Q=\left[ m'\left( {}_{6}{{C}^{11}} \right)-\left[ m'\left( {}_{5}^{11}B \right)+{{m}_{e}} \right] \right]{{c}^{2}}$…………………….. (1)

Where,${{m}_{e}}=$Mass of an electron or positron = \[0.000548u\]

$c=$ Speed of light

$m'=$Respective nuclear masses

If atomic masses are used instead of nuclear masses, then we will have to add $6{{m}_{e}}$ in the case of ${}^{11}C$ and $5{{m}_{e}}$in case of ${}^{11}B$.

Hence, equation (1) would now reduce to,

$\Delta Q=\left[ m\left( {}_{6}{{C}^{11}} \right)-m\left( {}_{5}^{11}B \right)-2{{m}_{e}} \right]{{c}^{2}}$

Where, $m\left( {}_{6}{{C}^{11}} \right)\text{ and m}\left( {}_{5}^{11}B \right)$are the atomic masses. 

Now, we have the change in Q value as, 

$\Delta Q=\left[ 11.011434-11.009305-2\times 0.000548 \right]{{c}^{2}}=\left( 0.001033{{c}^{2}} \right)u$

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

$\therefore \Delta Q=0.001033\times 931.5\approx 0.962MeV$

We see that the Q value is almost comparable to the maximum energy of the emitted positron. 

Ans: We know that: In ${{\beta }^{-}}$emission, the number of protons increases by 1, and one electron and an antineutrino are emitted from the parent nucleus. ${{\beta }^{-}}$emission of the nucleus could be given by, 

${}_{10}^{23}Ne\to {}_{11}^{23}Na+{{e}^{-}}+\bar{\nu }+Q$

It is also given that:

Atomic mass of ${}_{10}^{23}Ne=22.994466u$

Atomic mass of ${}_{11}^{23}Na=22.989770u$

Mass of an electron, ${{m}_{e}}=0.000548u$

Q value of the given reaction could be given as:

$Q=\left[ m\left( {}_{10}^{23}Ne \right)-\left[ m\left( {}_{11}^{23}Na \right)+{{m}_{e}} \right] \right]{{c}^{2}}$

There are 10 electrons in ${}_{10}N{{e}^{23}}$and 11 electrons in ${}_{11}^{23}Na$. Hence, the mass of the electron is cancelled in the Q-value equation.

$Q=\left[ 22.994466-22.9897770 \right]{{c}^{2}}=\left( 0.004696{{c}^{2}} \right)u$

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

$\Rightarrow Q=0.004696\times 931.5=4.374MeV$

The daughter nucleus is too heavy as compared to that of ${{e}^{-}}\text{ and }\bar{\nu }$. Hence, it carries negligible energy. The kinetic energy of the antineutrino is found to be nearly zero. 

Hence, the maximum kinetic energy of the emitted electrons is almost equal to the Q-value, i.e.,\[4.374MeV\].

Ans:The given nuclear reaction is:

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

Atomic mass of ${}_{1}^{1}H=1.007825u$

Atomic mass of ${}_{1}^{3}H=3.0164049u$

Atomic mass of ${}_{1}^{2}H=2.014102u$

According to the question, the Q-value of the reaction could be written as:

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

$\Rightarrow Q=\left[ 1.007825+3.016049-2\times 2.014102 \right]{{c}^{2}}=\left( -0.00433{{c}^{2}} \right)u$

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

$\therefore Q=-0.00433\times 931.5=-4.0334MeV$

The negative Q-value of this reaction shows that the given reaction is endothermic. 

1. ${}_{6}^{12}C+{}_{6}^{12}C\to {}_{10}^{20}Ne+{}_{2}^{4}He$

Ans: We are given that, 

Atomic mass of ${}_{6}^{12}C=12.0u$

Atomic mass of ${}_{10}^{12}Ne=19.992439u$

Atomic mass of ${}_{2}^{4}He=4.002603u$

The Q-value here could be given as, 

$Q=\left[ 2m\left( {}_{6}^{12}C \right)-m\left( {}_{10}^{20}Ne \right)-m\left( {}_{2}^{4}He \right) \right]{{c}^{2}}$

$\Rightarrow Q=\left[ 2\times 12.0-19.992439-4.002603 \right]{{c}^{2}}=\left( 0.004958{{c}^{2}} \right)u=0.004958\times 931.5$

$\therefore Q=4.618377MeV$

Since the Q-value is found to be positive, the reaction could be considered exothermic. 

Ans: We know that the fission of ${}_{26}^{56}Fe$ could be given as, 

${}_{26}^{56}Fe\to 2{}_{13}^{28}Al$

We are also given, atomic masses of ${}_{26}^{56}Fe\text{ and }{}_{13}^{28}Al$ as $55.93494u\text{ and 27}\text{.98191u}$ respectively. 

The Q-value here would be given as, 

$Q=\left[ m\left( {}_{26}^{56}Fe \right)-2m\left( {}_{13}^{28}Al \right) \right]{{c}^{2}}$

$\Rightarrow Q=\left[ 55.93494-2\times 27.98191 \right]{{c}^{2}}=\left( -0.02888{{c}^{2}} \right)u$

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

$\therefore Q=-0.02888\times 931.5=-26.902MeV$

The Q value is found to be negative and hence we could say that the fission is not possible energetically. In order for a reaction to be energetically possible, the Q-value must be positive. 

Ans: We are given that the average energy released per fission of ${}_{94}^{239}Pu$, ${{E}_{av}}=180MeV$

The amount of pure ${}_{94}P{{u}^{239}}$, $m=1kg=1000g$

Avogadro number, ${{N}_{A}}=6.023\times {{10}^{23}}$

Mass number of ${}_{94}^{239}Pu=239g$

1 mole of ${}_{94}P{{u}^{239}}$contains Avogadro number of atoms.

$1g\text{ of }{}_{94}P{{u}^{239}}\text{ contains }\left( \frac{{{N}_{A}}}{mass\text{ number}}\times m \right)\text{ atoms}$

$\Rightarrow \left( \frac{6.023\times {{10}^{23}}}{239}\times 1000 \right)=2.52\times {{10}^{24}}atoms$

Total energy released during the fission of 1kg of ${}_{94}^{239}Pu$could be calculated as:

$E={{E}_{av}}\times 2.52\times {{10}^{24}}=180\times 2.52\times {{10}^{24}}=4.536\times {{10}^{26}}MeV$

Therefore, $4.536\times {{10}^{26}}MeV$ is released if all the atoms in 1kg of pure ${}_{94}P{{u}^{239}}$undergo fission.

Ans: We are said that the half life of the fuel of the fission reactor, ${{t}_{\frac{1}{2}}}=5years$

$\Rightarrow {{t}_{\frac{1}{2}}}=\text{ }5\text{ }\times \text{ }365\text{ }\times \text{ }24\text{ }\times \text{ }60\text{ }\times \text{ }60\text{ }s$

We know that in the fission of 1g of ${}_{92}^{235}U$ nucleus, the energy released is equal to 200MeV.

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

$1g\text{ of }{}_{92}^{235}U\text{ contains }\frac{6.023\times {{10}^{23}}}{234}atoms$

The total energy generated per gram of ${}_{92}^{235}U$ is calculated as:

$E=\frac{6.023\times {{10}^{23}}}{235}\times 200MeV/g=\frac{200\times 6.023\times {{10}^{23}}\times 1.6\times {{10}^{-19}}\times {{10}^{6}}}{235}=8.20\times {{10}^{10}}J/g$

The reactor operator operates only 80% of the time. Therefore, the amount of ${}_{92}^{235}U$ consumed in 5years by the 1000MW fission reactor could be calculated as, 

$\frac{5\times 80\times 60\times 60\times 365\times 24\times 1000\times {{10}^{6}}}{100\times 8.20\times {{10}^{10}}}g\approx 1538kg$

So, the initial amount of ${}_{92}^{235}U=2\times 1538=3076kg$

Hence, we found the initial amount of uranium to be $3076kg.$

Ans: The fusion reaction is given to be:

${}_{1}^{2}H+{}_{1}^{2}H\to {}_{2}^{3}He+n+3.27MeV$

Amount of deuterium, \[m=2\text{ }kg\]

1 mole, i.e., 2 g of deuterium contains \[6.023\text{ }\times \text{ }{{10}^{23}}atoms\].

2.0 kg of deuterium contains $\frac{6.023\times {{10}^{23}}}{2}\times 2000=6.023\times {{10}^{26}}atoms$ atoms

It could be inferred from the given reaction that when two atoms of deuterium fuse, \[3.27MeV\]energy is released.

Therefore, the total energy per nucleus released in the fusion reaction would be:

$E=\frac{3.27}{2}\times 6.023\times {{10}^{26}}MeV=\frac{3.27}{2}\times 6.023\times {{10}^{26}}\times 1.6\times {{10}^{-19}}\times {{10}^{6}}$

$\therefore E=1.576\times {{10}^{14}}J$

Power of the electric lamp is given to be, $P=100W=100J/s$, that is, the energy consumed by the lamp per second is 100J. 

Now, the total time for which the electric lamp glows could be calculated as, 

$t=\frac{1.576\times {{10}^{14}}}{100}=\frac{1.576\times {{10}^{14}}}{100\times 60\times 60\times 24\times 365}$

\[\therefore t\approx 4.9\times {{10}^{4}}years\]

Hence, the total time for which the electric lamp glows is found to be $4.9\times {{10}^{4}}years.$

Ans: When two deuterons collide head-on, the distance between their centres, d could be given as:

Radius of 1st deuteron $+$Radius of 2nd deuteron

Radius of a deuteron nucleus$=2fm=2\times {{10}^{-15}}m$

$\Rightarrow d=2\times {{10}^{-15}}+2\times {{10}^{-15}}=4\times {{10}^{-15}}m$

Also, charge on a deuteron $=$ Charge on an electron $=e=1.6\times {{10}^{-19}}C$

Potential energy of the two-deuteron system could be given by, 

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

Where, ${{\varepsilon }_{0}}$is the permittivity of free space. 

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

$\Rightarrow V=\frac{9\times {{10}^{9}}\times {{\left( 1.6\times {{10}^{19}} \right)}^{2}}}{4\times {{10}^{15}}}J=\frac{9\times {{10}^{9}}\times {{\left( 1.6\times {{10}^{-19}} \right)}^{2}}}{4\times {{10}^{-15}}\times \left( 1.6\times {{10}^{-19}} \right)}eV$

$\therefore V=360keV$

Therefore, we found the height of the potential barrier of the two-deuteron system to be 360keV.