Nuclei
Exercises
You may find the following data useful in solving the exercises:
$$ \begin{aligned} &e=1.6 \times 10^{-19}\text{C} & & N= 6.023 \times 10 ^{23} \text{per mole}\\ &\frac{1}{(4 \pi \varepsilon _0)}=9 \times 10 ^9 \text{N} m ^2/C ^2 && k=1.381 \times 10 ^{-23} \text{J} k ^{-1} \\ &1 \text{MeV}=1.6 \times 10 ^{-13} \text{J} && 1 \text{u} = 931.5 \text{MeV}/c ^2 \\ &1 \text{ year} = 3.154 \times 10 ^7 \text{s} \\ & \text{m}_H=4.002603 \text{ u} && \text{m}_n=1.007825 \text{u} \\ & m( ^4_2\text{He})=4.002603 u && \text{m}_e=0.000548 \text{u} \end{aligned} $$
13.1 Obtain the binding energy (in $\mathrm{MeV}$ ) of a nitrogen nucleus $( _{7} ^{14} \mathrm{~N})$, given $m( _{7} ^{14} \mathrm{~N})=14.00307 \mathrm{u}$
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Answer
Atomic mass of nitrogen $({ }_{7} \mathrm{~N} ^{14}), m=14.00307 \mathrm{u}$
A nucleus of nitrogen ${ }_{7} \mathrm{~N} ^{14}$ contains 7 protons and 7 neutrons.
Hence, the mass defect of this nucleus, $\Delta m=7 m_{H}+7 m_{n}-m$
Where,
Mass of a proton, $m_{H}=1.007825 \mathrm{u}$
Mass of a neutron, $m_{n}=1.008665 \mathrm{u}$
$\therefore \Delta m=7 \times 1.007825+7 \times 1.008665-14.00307$ $=7.054775+7.06055-14.00307$
$=0.11236 \mathrm{u}$
But $1 \mathrm{u}=931.5 \mathrm{MeV} / \mathrm{c} ^{2}$
$\therefore \Delta m=0.11236 \times 931.5 \mathrm{MeV} / c ^{2}$
Hence, the binding energy of the nucleus is given as:
$E_{b}=\Delta m c ^{2}$
Where,
$c=$ Speed of light
$\therefore E_{b}=0.11236 \times 931.5(\frac{\mathrm{MeV}}{c ^{2}}) \times c ^{2}$
$=104.66334 \mathrm{MeV}$
Hence, the binding energy of a nitrogen nucleus is $104.66334 \mathrm{MeV}$.
13.2 Obtain the binding energy of the nuclei $ _{26} ^{56} \mathrm{Fe}$ and $ _{83} ^{209} \mathrm{Bi}$ in units of $\mathrm{MeV}$ from the following data:
$$ m( _{26} ^{56} \mathrm{Fe})=55.934939 \mathrm{u} \quad m( _{83} ^{209} \mathrm{Bi})=208.980388 \mathrm{u} $$
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Answer
Atomic mass of ${ } _{26} ^{56} \mathrm{Fe}, m _{1}=55.934939 \mathrm{u}$
${ } _{26} ^{56} \mathrm{Fe}$ nucleus has 26 protons and $(56-26)=30$ neutrons
Hence, the mass defect of the nucleus, $\Delta m=26 \times m _{H}+30 \times m _{n}-m _{1}$
Where,
Mass of a proton, $m _{H}=1.007825 \mathrm{u}$
Mass of a neutron, $m _{n}=1.008665 \mathrm{u}$
$\therefore \Delta m=26 \times 1.007825+30 \times 1.008665-55.934939$
$=26.20345+30.25995-55.934939$
$=0.528461 \mathrm{u}$
But $1 \mathrm{u}=931.5 \mathrm{MeV} / \mathrm{c} ^{2}$
$\therefore \Delta m=0.528461 \times 931.5 \mathrm{MeV} / c ^{2}$
The binding energy of this nucleus is given as:
$E _{b 1}=\Delta m c ^{2}$
Where,
$c=$ Speed of light
$\therefore E _{b 1}=0.528461 \times 931.5(\frac{\mathrm{MeV}}{c ^{2}}) \times c ^{2}$
$=492.26 \mathrm{MeV}$
Average binding energy per nucleon $=\frac{492.26}{56}=8.79 \mathrm{MeV}$
Atomic mass of ${ } ^{\frac{209}{83} \mathrm{Bi}}, m _{2}=208.980388 \mathrm{u}$
${ } _{83} ^{2099} \mathrm{Bi}$ nucleus has 83 protons and $(209-83) 126$ neutrons.
Hence, the mass defect of this nucleus is given as:
$\Delta m ^{\prime}=83 \times m _{H}+126 \times m _{n}-m _{2}$
Where,
Mass of a proton, $m _{H}=1.007825 \mathrm{u}$
Mass of a neutron, $m _{n}=1.008665 \mathrm{u}$
$\therefore \Delta m ^{\prime}=83 \times 1.007825+126 \times 1.008665-208.980388$
$=83.649475+127.091790-208.980388$ $=1.760877 \mathrm{u}$
But $1 \mathrm{u}=931.5 \mathrm{MeV} / \mathrm{c} ^{2}$
$\therefore \Delta m ^{\prime}=1.760877 \times 931.5 \mathrm{MeV} / c ^{2}$
Hence, the binding energy of this nucleus is given as:
$E _{b 2}=\Delta m ^{\prime} c ^{2}$
$=1.760877 \times 931.5(\frac{\mathrm{MeV}}{c ^{2}}) \times c ^{2}$
$=1640.26 \mathrm{MeV}$
Average bindingenergy per nucleon $=\frac{1640.26}{209}=7.848 \mathrm{MeV}$
13.3 A given coin has a mass of $3.0 \mathrm{~g}$. Calculate the nuclear energy that would be required to separate all the neutrons and protons from each other. For simplicity assume that the coin is entirely made of $ _{29} ^{63} \mathrm{Cu}$ atoms (of mass $62.92960 \mathrm{u}$ ).
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Answer
Mass of a copper coin, $m ^{\prime}=3 \mathrm{~g}$
Atomic mass of ${ } _{29} \mathrm{Cu} ^{63}$ atom, $m=62.92960 \mathrm{u}$
The total number of ${ } _{29} \mathrm{Cu} ^{63}$ atoms in the coin,$N=\frac{N _{\mathrm{A}} \times m ^{\prime}}{\text { Mass number }}$
Where,
$\mathrm{N} _{\mathrm{A}}=$ Avogadro’s number $=6.023 \times 10 ^{23}$ atoms $/ \mathrm{g}$
Mass number $=63 \mathrm{~g}$ $\therefore N=\frac{6.023 \times 10 ^{23} \times 3}{63}=2.868 \times 10 ^{22}$ atoms
${ } _{29} \mathrm{Cu} ^{63}$ nucleus has 29 protons and $(63-29) 34$ neutrons
$\therefore$ Mass defect of this nucleus, $\Delta m ^{\prime}=29 \times m _{H}+34 \times m _{n}-m$
Where,
Mass of a proton, $m _{H}=1.007825 \mathrm{u}$
Mass of a neutron, $m _{n}=1.008665 \mathrm{u}$
$\therefore \Delta m ^{\prime}=29 \times 1.007825+34 \times 1.008665-62.9296$
$=0.591935 \mathrm{u}$
Mass defect of all the atoms present in the coin, $\Delta m=0.591935 \times 2.868 \times 10 ^{22}$
$=1.69766958 \times 10 ^{22} \mathrm{u}$
But $1 \mathrm{u}=931.5 \mathrm{MeV} / \mathrm{c} ^{2}$
$\therefore \Delta m=1.69766958 \times 10 ^{22} \times 931.5 \mathrm{MeV} / c ^{2}$
Hence, the binding energy of the nuclei of the coin is given as:
$E _{b}=\Delta m c ^{2}$
$=1.69766958 \times 10 ^{22} \times 931.5(\frac{\mathrm{MeV}}{c ^{2}}) \times c ^{2}$
$=1.581 \times 10 ^{25} \mathrm{MeV}$
But $1 \mathrm{MeV}=1.6 \times 10 ^{-13} \mathrm{~J}$
$E _{b}=1.581 \times 10 ^{25} \times 1.6 \times 10 ^{-13}$
$=2.5296 \times 10 ^{12} \mathrm{~J}$
This much energy is required to separate all the neutrons and protons from the given coin.
13.4 Obtain approximately the ratio of the nuclear radii of the gold isotope $ _{79} ^{197} \mathrm{Au}$ and the silver isotope $ _{47} ^{107} \mathrm{Ag}$.
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Answer
Nuclear radius of the gold isotope ${ } _{79} \mathrm{Au} ^{197}=R _{\mathrm{Au}}$
Nuclear radius of the silver isotope ${ } _{47} \mathrm{Ag} ^{107}=R _{\mathrm{Ag}}$
Mass number of gold, $A _{\mathrm{Au}}=197$
Mass number of silver, $A _{\mathrm{Ag}}=107$
The ratio of the radii of the two nuclei is related with their mass numbers as:
$$ \begin{aligned} \frac{R _{\mathrm{Au}}}{R _{\mathrm{Ag}}} & =(\frac{R _{\mathrm{Au}}}{R _{\mathrm{Ag}}}) ^{\frac{1}{3}} \ & =(\frac{197}{107}) ^{\frac{1}{3}}=1.2256 \end{aligned} $$
Hence, the ratio of the nuclear radii of the gold and silver isotopes is about 1.23.
13.5 The $Q$ value of a nuclear reaction $A+b \rightarrow C+d$ is defined by
$Q=\left[m _{A}+m _{b}-m _{C}-m _{d}\right] c ^{2}$
where the masses refer to the respective nuclei. Determine from the given data the $Q$-value of the following reactions and state whether the reactions are exothermic or endothermic.
(i) $ _{1} ^{1} \mathrm{H}+ _{1} ^{3} \mathrm{H} \rightarrow _{1} ^{2} \mathrm{H}+ _{1} ^{2} \mathrm{H}$
(ii) $ _{6} ^{12} \mathrm{C}+ _{6} ^{12} \mathrm{C} \rightarrow _{10} ^{20} \mathrm{Ne}+ _{2} ^{4} \mathrm{He}$
Atomic masses are given to be
$m( _{1} ^{2} \mathrm{H})=2.014102 \mathrm{u}$
$m( _{1} ^{3} \mathrm{H})=3.016049 \mathrm{u}$
$m( _{6} ^{12} \mathrm{C})=12.000000 \mathrm{u}$
$m( _{10} ^{20} \mathrm{Ne})=19.992439 \mathrm{u}$
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Answer
Alpha particle decay of ${ } ^{226} \mathrm{Ra}$ emits a helium nucleus. As a result, its mass number reduces to $(226-4) 222$ and its atomic number reduces to $(88-2) 86$. This is shown in the following nuclear reaction.
${ } _{88} ^{226} \mathrm{Ra} \longrightarrow{ } _{86} ^{222} \mathrm{Ra}+{ } _{2} ^{4} \mathrm{He}$
$Q$-value of
emitted $\alpha$-particle $=($ Sum of initial mass - Sum of final mass $) c ^{2}$
Where,
$c=$ Speed of light
It is given that:
$m({ } _{88} ^{226} \mathrm{Ra})=226.02540 \mathrm{u}$
$m({ } _{86} ^{222} \mathrm{Rn})=222.01750 \mathrm{u}$
$m({ } _{2} ^{4} \mathrm{He})=4.002603 \mathrm{u}$
$Q$-value $=[226.02540-(222.01750+4.002603)] \mathrm{u} c ^{2}$
$=0.005297 \mathrm{u} c ^{2}$
But $1 \mathrm{u}=931.5 \mathrm{MeV} / \mathrm{c} ^{2}$
$\therefore Q=0.005297 \times 931.5 \approx 4.94 \mathrm{MeV}$
Kinetic energy of the $\alpha$-particle $=(\frac{\text { Mass number after decay }}{\text { Mass number before decay }}) \times Q$
$=\frac{222}{226} \times 4.94=4.85 \mathrm{MeV}$
Alpha particle decay of $({ } _{86} ^{220} \mathrm{Rn})$ is shown by the following nuclear reaction.
${ } _{86} ^{220} \mathrm{Rn} \longrightarrow{ } _{84} ^{216} \mathrm{Po}+{ } _{2} ^{4} \mathrm{He}$
It is given that:
Mass of $({ } _{86} ^{220} \mathrm{Rn})=220.01137 \mathrm{u}$
Mass of $({ } ^{24}{ } ^{216} \mathrm{Po})=216.00189 \mathrm{u}$
$\therefore Q$-value $=[220.01137-(216.00189+4.00260)] \times 931.5$
$\approx 641 \mathrm{MeV}$
Kinetic energy of the $\alpha$-particle $=(\frac{220-4}{220}) \times 6.41$
$=6.29 \mathrm{MeV}$
13.6 Suppose, we think of fission of a $ _{26} ^{56} \mathrm{Fe}$ nucleus into two equal fragments, $ _{13} ^{28} \mathrm{Al}$. Is the fission energetically possible? Argue by working out $Q$ of the process. Given $m( _{26} ^{56} \mathrm{Fe})=55.93494 \mathrm{u}$ and $m( _{13} ^{28} \mathrm{Al})=27.98191$ u.
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Answer
The fission of ${ } _{26} ^{56} \mathrm{Fe}$ can be given as:
$$ { } _{13} ^{56} \mathrm{Fe} \longrightarrow 2{ } _{13} ^{28} \mathrm{Al} $$
It is given that:
Atomic mass of $m({ } _{26} ^{56} \mathrm{Fe})=55.93494 \mathrm{u}$
Atomic mass of $m({ } _{13} ^{28} \mathrm{Al})=27.98191 \mathrm{u}$
The $Q$-value of this nuclear reaction is given as:
$$ \begin{aligned} Q & =\left[m({ } _{26} ^{56} \mathrm{Fe})-2 m({ } _{13} ^{28} \mathrm{Al})\right] c ^{2} \ & =[55.93494-2 \times 27.98191] c ^{2} \ & =(-0.02888 c ^{2}) \mathrm{u} \end{aligned} $$
But $1 \mathrm{u}=931.5 \mathrm{MeV} / \mathrm{c} ^{2}$
$\therefore Q=-0.02888 \times 931.5=-26.902 \mathrm{MeV}$
The $Q$-value of the fission is negative. Therefore, the fission is not possible energetically. For an energetically-possible fission reaction, the $Q$-value must be positive.
13.7 The fission properties of $ _{94} ^{239} \mathrm{Pu}$ are very similar to those of $ _{92} ^{235} \mathrm{U}$. The average energy released per fission is $180 \mathrm{MeV}$. How much energy, in $\mathrm{MeV}$, is released if all the atoms in $1 \mathrm{~kg}$ of pure $ _{94} ^{239} \mathrm{Pu}$ undergo fission?
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Answer
Average energy released per fission of ${ } _{94} ^{239} \mathrm{Pu}, E _{a v}=180 \mathrm{MeV}$
Amount of pure ${ } _{94} \mathrm{Pu} ^{239}, m=1 \mathrm{~kg}=1000 \mathrm{~g}$
$\mathrm{N} _{\mathrm{A}}=$ Avogadro number $=6.023 \times 10 ^{23}$
Mass number of ${ } _{94} ^{239} \mathrm{Pu}=239 \mathrm{~g}$
1 mole of ${ } _{94} \mathrm{Pu} ^{239}$ contains $\mathrm{N} _{\mathrm{A}}$ atoms.
$\therefore m$ g of ${ } _{94} \mathrm{Pu} ^{239}$ contains $(\frac{\mathrm{N} _{\mathrm{A}}}{\text { Mass number }} \times m)$ atoms
$=\frac{6.023 \times 10 ^{23}}{239} \times 1000=2.52 \times 10 ^{24}$ atoms
$\therefore$ Total energy released during the fission of $1 \mathrm{~kg}$ of ${ } _{94} ^{239} \mathrm{Pu}$ is calculated as:
$$ \begin{aligned} E & =E _{\alpha v} \times 2.52 \times 10 ^{24} \ & =180 \times 2.52 \times 10 ^{24}=4.536 \times 10 ^{26} \mathrm{MeV} \end{aligned} $$
Hence, $4.536 \times 10 ^{26} \mathrm{MeV}$ is released if all the atoms in $1 \mathrm{~kg}$ of pure ${ } _{94} \mathrm{Pu} ^{239}$ undergo fission.
13.8 How long can an electric lamp of 100W be kept glowing by fusion of $2.0 \mathrm{~kg}$ of deuterium? Take the fusion reaction as
$$ _{1} ^{2} \mathrm{H}+ _{1} ^{2} \mathrm{H} \rightarrow _{2} ^{3} \mathrm{He}+\mathrm{n}+3.27 \mathrm{MeV} $$
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Answer
The given fusion reaction is:
${ } _{1} ^{2} \mathrm{H}+{ } _{1} ^{2} \mathrm{H} \longrightarrow{ } _{2} ^{3} \mathrm{He}+\mathrm{n}+3.27 \mathrm{MeV}$
Amount of deuterium, $m=2 \mathrm{~kg}$
1 mole, i.e., $2 \mathrm{~g}$ of deuterium contains $6.023 \times 10 ^{23}$ atoms.
$\therefore 2.0 \mathrm{~kg}$ of deuterium contains $=\frac{6.023 \times 10 ^{23}}{2} \times 2000=6.023 \times 10 ^{26}$ atoms
It can be inferred from the given reaction that when two atoms of deuterium fuse, 3.27 $\mathrm{MeV}$ energy is released.
$\therefore$ Total energy per nucleus released in the fusion reaction:
$$ \begin{aligned} E & =\frac{3.27}{2} \times 6.023 \times 10 ^{26} \mathrm{MeV} \ & =\frac{3.27}{2} \times 6.023 \times 10 ^{26} \times 1.6 \times 10 ^{-19} \times 10 ^{6} \ & =1.576 \times 10 ^{14} \mathrm{~J} \end{aligned} $$
Power of the electric lamp, $P=100 \mathrm{~W}=100 \mathrm{~J} / \mathrm{s}$
Hence, the energy consumed by the lamp per second $=100 \mathrm{~J}$
The total time for which the electric lamp will glow is calculated as:
$$ \begin{aligned} & \frac{1.576 \times 10 ^{14}}{100} \mathrm{~s} \ & \frac{1.576 \times 10 ^{14}}{100 \times 60 \times 60 \times 24 \times 365} \approx 4.9 \times 10 ^{4} \text { years } \end{aligned} $$
13.9 Calculate the height of the potential barrier for a head on collision of two deuterons. (Hint: The height of the potential barrier is given by the Coulomb repulsion between the two deuterons when they just touch each other. Assume that they can be taken as hard spheres of radius $2.0 \mathrm{fm}$.)
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Answer
When two deuterons collide head-on, the distance between their centres, $d$ is given as:
Radius of $1 ^{\text {st }}$ deuteron + Radius of $2 ^{\text {nd }}$ deuteron
Radius of a deuteron nucleus $=2 \mathrm{fm}=2 \times 10 ^{-15} \mathrm{~m}$
$\therefore d=2 \times 10 ^{-15}+2 \times 10 ^{-15}=4 \times 10 ^{-15} \mathrm{~m}$
Charge on a deuteron nucleus $=$ Charge on an electron $=e=1.6 \times 10 ^{-19} \mathrm{C}$
Potential energy of the two-deuteron system:
$$ V=\frac{e ^{2}}{4 \pi \epsilon _{0} d} $$
Where,
$$ \epsilon _{0}=\text { Permittivity of free space } $$
$$ \frac{1}{4 \pi \epsilon _{0}}=9 \times 10 ^{9} \mathrm{~N} \mathrm{~m} ^{2} \mathrm{C} ^{-2} $$
$\therefore V=\frac{9 \times 10 ^{9} \times(1.6 \times 10 ^{-19}) ^{2}}{4 \times 10 ^{-15}} \mathrm{~J}$
$$ =\frac{9 \times 10 ^{9} \times(1.6 \times 10 ^{-19}) ^{2}}{4 \times 10 ^{-15} \times(1.6 \times 10 ^{-19})} \mathrm{eV} $$
$$ =360 \mathrm{keV} $$
Hence, the height of the potential barrier of the two-deuteron system is
$360 \mathrm{keV}$.
13.10 From the relation $R=R _{0} A ^{1 / 3}$, where $R _{0}$ is a constant and $A$ is the mass number of a nucleus, show that the nuclear matter density is nearly constant (i.e. independent of $A$ ).
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Answer
We have the expression for nuclear radius as:
$R=R _{0} A ^{1} \beta ^{3}$
Where,
$R _{0}=$ Constant.
$A=$ Mass number of the nucleus
Nuclear matter density, $\rho=\frac{\text { Mass of the nucleus }}{\text { Volume of the nucleus }}$
Let $m$ be the average mass of the nucleus.
Hence, mass of the nucleus $=m A$
$\therefore \rho=\frac{m A}{\frac{4}{3} \pi R ^{3}}=\frac{3 m A}{4 \pi(R _{0} A ^{\frac{1}{3}}) ^{3}}=\frac{3 m A}{4 \pi R _{0} ^{3} A}=\frac{3 m}{4 \pi R _{0} ^{3}}$
Hence, the nuclear matter density is independent of $A$. It is nearly constant.
