Chapter 13: Nuclei

Ans: We are given the following information:

Mass of ${}_{3}^{6}Li$ lithium isotope, ${{m}_{1}}=6.01512u$

Mass of ${}_{3}^{7}Li$ lithium isotope, ${{m}_{2}}=7.01600u$

Abundance of${}_{3}^{6}Li$, ${{n}_{1}}=7.5%$

Abundance of  ${}_{3}^{7}Li$, ${{n}_{2}}=92.5%$

The atomic mass of lithium atom is given by,

$m=\frac{{{m}_{1}}{{n}_{1}}+{{m}_{2}}{{n}_{2}}}{{{n}_{1}}+{{n}_{2}}}$

Substituting the given values, we get,

$m=\frac{6.01512\times 7.5+7.01600\times 92.5}{92.5+7.5}$

$\therefore m=6.940934u$

Therefore, we found the atomic mass of lithium atoms to be $6.940934u$. 

Ans: We are given:

Mass of ${}_{5}^{10}B$ Boron isotope, ${{m}_{1}}=10.01294u$

Mass of ${}_{5}^{11}B$ lithium isotope, ${{m}_{2}}=11.00931u$

Abundance of${}_{5}^{10}B$, ${{n}_{1}}=x%$

Abundance of  ${}_{5}^{11}B$, ${{n}_{2}}=\left( 100-x \right)%$

We know the atomic mass of boron to be, $m=10.811u$

The atomic mass of lithium atom is given by,

$m=\frac{{{m}_{1}}{{n}_{1}}+{{m}_{2}}{{n}_{2}}}{{{n}_{1}}+{{n}_{2}}}$

Substituting the given values, we get,

$10.811=\frac{10.01294\times x+11.00931\times \left( 100-x \right)}{x+\left( 100-x \right)}$

$\Rightarrow 1081.11=10.01294x+1100.931-11.00931x$

$\therefore x=\frac{19.821}{0.99637}=19.89%$

And, $100-x=80.11%$

Therefore, we found the abundance of ${}_{5}^{10}B$and ${}_{5}^{11}B$to be $19.89%$ and $80.11%$respectively. 

Ans: We are given that:

Atomic mass of  ${}_{10}^{20}Ne$, ${{m}_{1}}=19.99u$

Abundance of ${}_{10}^{20}Ne$, ${{\eta }_{1}}=90.51%$

Atomic mass of${}_{10}^{21}Ne$, ${{m}_{2}}=20.99u$

Abundance of ${}_{10}^{21}Ne$, ${{\eta }_{2}}=0.27%$

Atomic mass of ${}_{10}^{22}Ne$, ${{m}_{3}}=21.99u$

Abundance of ${}_{10}^{22}Ne$, ${{\eta }_{3}}=9.22%$

The average atomic mass of neon could be given as,

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

Substituting the given values, we get, 

$m=\frac{19.99\times 90.51+20.99\times 0.27+21.99\times 9.22}{90.51+0.27+9.22}$

$\therefore m=20.1771u$

The average atomic mass of neon is thus found to be $20.177u.$

Ans: We are given:

Atomic mass of nitrogen $\left( {}_{7}{{N}^{14}} \right)$, $m=14.00307u$

A nucleus of ${}_{7}{{N}^{14}}$nitrogen contains 7 protons and 7 neutrons.

Hence, the mass defect of this nucleus would be, $\Delta m=7{{m}_{H}}+7{{m}_{n}}-m$

Where,

Mass of a proton, ${{m}_{H}}=1.007825u$

Mass of a neutron, ${{m}_{n}}=1.008665u$

Substituting these values into the above equation, we get, 

$\Delta m=7\times 1.007825+7\times 1.008665-14.00307$

$\Rightarrow \Delta m=7.054775+7.06055-14.00307$

$\therefore \Delta m=0.11236u$

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

$\Rightarrow \Delta m=0.11236\times 931.5MeV/{{c}^{2}}$

Now, we could give the binding energy as, 

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

Where, $c=\text{speed of light =3}\times \text{1}{{\text{0}}^{8}}m{{s}^{-2}}$

Now, ${{E}_{b}}=0.11236\times 931.5\left( \frac{MeV}{{{c}^{2}}} \right)\times {{c}^{2}}$

$\therefore {{E}_{b}}=104.66334MeV$

Therefore, we found the binding energy of a Nitrogen nucleus to be $104.66334MeV$. 

Ans: We are given the following:

Atomic mass of ${}_{26}^{56}Fe$, ${{m}_{1}}=55.934939u$

${}_{26}^{56}Fe$ nucleus has 26 protons and $56-26=30$neutrons

Hence, the mass defect of the nucleus would be, $\Delta m=26\times {{m}_{H}}+30\times {{m}_{n}}-{{m}_{1}}$

Where, Mass of a proton,${{m}_{H}}=1.007825u$

Mass of a neutron,${{m}_{n}}=1.008665u$

Substituting these values into the above equation, we get, 

$\Delta m=26\times 1.007825+30\times 1.008665-55.934939$

$\Rightarrow \Delta m=26.20345+30.25995-55.934939$

$\therefore \Delta m=0.528461u$

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

$\Delta m=0.528461\times 931.5MeV/{{c}^{2}}$

The binding energy of this nucleus could be given as,

${{E}_{b1}}=\Delta m{{c}^{2}}$

Where, c = Speed of light

$\Rightarrow {{E}_{b1}}=0.528461\times 931.5\left( \frac{MeV}{{{c}^{2}}} \right)\times {{c}^{2}}$

$\therefore {{E}_{b1}}=492.26MeV$

Now, we have the average binding energy per nucleon to be, 

$B.E=\frac{492.26}{56}=8.79MeV$

Also, atomic mass of ${}_{83}^{209}Bi$, ${{m}_{2}}=208.980388u$

We know that, ${}_{83}^{209}Bi$ nucleus has 83 protons and $209-83=126\text{ neutrons}$

Where,

Mass of a proton, ${{m}_{H}}=1.007825u$

Mass of a neutron, ${{m}_{n}}=1.008665u$

$\Delta m'=83\times 1.007825+126\times 1.008665-208.980388$

$\Rightarrow \Delta m'=83.649475+127.091790-208.980388$

$\therefore \Delta m'=1.760877u$

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

Hence, the binding energy of this nucleus could be given as,

${{E}_{b2}}=\Delta m'{{c}^{2}}=1.760877\times 931.5\left( \frac{MeV}{{{c}^{2}}} \right)\times {{c}^{2}}$

$\therefore {{E}_{b2}}=1640.26MeV$

Average binding energy per nucleon is found to be $=\frac{1640.26}{209}=7.848MeV$

Hence, the average binding energy per nucleon is found to be $7.848MeV$. 

Ans: We are given:

Mass of a copper coin, $m'=3g$

Atomic mass of ${}_{29}C{{u}^{63}}$atom, \[m=62.92960u\]

The total number of ${}_{29}^{63}Cu$ atoms in the coin, $N=\frac{{{N}_{A}}\times m'}{\text{Mass number}}$

Where,\[{{N}_{A}}=\text{ }Avogadro\text{ }number=6.023\times \text{ }{{10}^{23}}atoms/g\]

Mass number = 63g

$\Rightarrow N=\frac{6.023\times {{10}^{23}}\times 3}{63}=2.868\times {{10}^{22}}$

${}_{29}C{{u}^{63}}$nucleus has 29 protons and \[\left( 63-29 \right)=34\]neutrons

Mass defect of this nucleus would be, $\Delta m'=29\times {{m}_{H}}+34\times {{m}_{n}}-m$

Where,

Mass of a proton, ${{m}_{H}}=1.007825u$

Mass of a neutron, ${{m}_{n}}=1.008665u$$\Delta m'=29\times 1.007825+34\times 1.008665-62.9296=0.591935u$

Mass defect of all the atoms present in the coin would be, 

$\Delta m=0.591935\times 2.868\times {{10}^{22}}=1.69766958\times {{10}^{22}}u$

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

$\Rightarrow \Delta m=1.69766958\times {{10}^{22}}\times 931.5MeV/{{c}^{2}}$

Hence, the binding energy of the nuclei of the coin could be given as:

${{E}_{b}}=\Delta m{{c}^{2}}=1.69766958\times {{10}^{22}}\times 931.5\left( \frac{MeV}{{{c}^{2}}} \right)\times {{c}^{2}}$

$\therefore {{E}_{b}}=1.581\times {{10}^{25}}MeV$

But, $1MeV=1.6\times {{10}^{-13}}J$

$\Rightarrow {{E}_{b}}=1.581\times {{10}^{25}}\times 1.6\times {{10}^{-13}}$

$\therefore {{E}_{b}}=2.5296\times {{10}^{12}}J$

This much energy is needed to separate all the neutrons and protons from the given coin.

Ans: We know that, $\alpha $is basically a nucleus of Helium $\left( {}_{2}H{{e}^{4}} \right)$and $\beta $is an electron $\left( {{e}^{-}}\text{ for }{{\beta }^{-}}\text{ and }{{\text{e}}^{+}}\text{ for }{{\beta }^{+}} \right)$. In every $\alpha $-decay, there is a loss of 2 protons and 2 neutrons. In every ${{\beta }^{+}}$-decay, there is a loss of 1 proton and a neutrino is emitted from the nucleus. In every ${{\beta }^{-}}$-decay, there is a gain of 1 proton and an antineutrino is emitted from the nucleus.

For the given case, the nuclear reaction would be, 

${}_{88}R{{a}^{226}}\to {}_{86}R{{n}^{222}}+{}_{2}H{{e}^{4}}$

Ans: ${}_{94}^{242}Pu\to {}_{92}^{238}U+{}_{2}^{4}He$

1. $\beta -\text{decay of }{}_{15}^{32}P$

Ans: ${}_{15}^{32}P\to {}_{16}^{32}S+{{e}^{-}}+\bar{\nu }$

1. $\beta -\text{decay of }{}_{83}^{210}Bi$

Ans: ${}_{83}^{210}B\to {}_{84}^{210}Po+{{e}^{-}}+\bar{\nu }$

1. ${{\beta }^{+}}-\text{decay of }{}_{6}^{11}C$

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

1. ${{\beta }^{+}}-\text{decay of }{}_{43}^{97}Tc$

Ans: ${}_{43}^{97}Tc\to {}_{42}^{97}Mo+{{e}^{+}}+\nu $

1. Electron capture of ${}_{54}^{120}Xe$

Ans: ${}_{54}^{120}Xe+{{e}^{+}}\to {}_{53}^{120}I+\nu $

Ans: We are said that, Half-life of the radioactive isotope$=\text{ T years}$

Original amount of the radioactive isotope $={{N}_{0}}$

(a) After decay, let the amount of the radioactive isotope be N. 

It is given that only $3.125%$ of ${{N}_{0}}$remains after decay. Hence, we could write,

$\frac{N}{{{N}_{0}}}=3.125%=\frac{3.125}{100}=\frac{1}{32}$

But we know that, $\frac{N}{{{N}_{0}}}={{e}^{-\lambda t}}$

Where, $\lambda =$Decay constant and $t=$Time

$\Rightarrow -\lambda t=\frac{1}{32}$

$\Rightarrow -\lambda t=\ln l-\ln 32$

$\Rightarrow -\lambda t=0-3.4657$

$\Rightarrow t=\frac{3.4657}{\lambda }$

But, since $\lambda =\frac{0.693}{T}$

$\Rightarrow t=\frac{3.466}{\left( \frac{0.693}{T} \right)}$

$\therefore t\approx 5T\text{ years}$

Therefore, we found that the isotope will take about 5T years in order to reduce to $3.125%$ of its original value. 

1. 1% of its original value?

Ans: After decay, let the amount of the radioactive isotope be N

It is given that only 1% of ${{N}_{0}}$ remains after decay. Hence, we could write:

$\frac{N}{{{N}_{0}}}=1%=\frac{1}{100}$

But we know, $\frac{N}{{{N}_{0}}}={{e}^{-\lambda t}}$

$\Rightarrow {{e}^{-\lambda t}}=\frac{1}{100}$

$\Rightarrow -\lambda t=\ln 1-\ln 100$

$\Rightarrow -\lambda t=0-4.602$

$\Rightarrow t=\frac{4.6052}{\lambda }$

Since we have, $\lambda =\frac{0.639}{T}$

$\Rightarrow t=\frac{4.6052}{\left( \frac{0.693}{T} \right)}$

$\therefore t=6.645T\text{ years}$

Therefore, we found that the given isotope would take about 6.645T years to reduce to 1% of its original value. 

Ans: We are given that: 

Decay rate of living carbon-containing matter, \[R=15\text{ }decay/min\]

Let N be the number of radioactive atoms present in a normal carbon- containing matter.

Half life of ${}_{6}^{14}C,\text{ }{{\text{T}}_{\frac{1}{2}}}=5730\text{ years}$

The decay rate of the specimen obtained from the Mohenjodaro site:

\[R'=9\text{ }decays/min\]

Let N' be the number of radioactive atoms present in the specimen during the Mohenjodaro period.

Therefore, we can relate the decay constant, $\lambda $ and time, t as:

$\frac{N'}{N}=\frac{R'}{R}={{e}^{-\lambda t}}$

$\Rightarrow {{e}^{-\lambda t}}=\frac{9}{15}=\frac{3}{5}$

$\Rightarrow -\lambda t={{\log }_{e}}\frac{3}{5}=-0.5108$

$\Rightarrow t=\frac{0.5108}{\lambda }$

But we know, 

$\lambda =\frac{0.693}{{{T}_{\frac{1}{2}}}}=\frac{0.693}{5730}$

$\Rightarrow t=\frac{0.5108}{\left( \frac{0.693}{5730} \right)}=4223.5\text{ years}$

Therefore, the approximate age of the Indus-Valley civilization is found to be 4223.5 years.

Ans: We know that, 

The strength of the radioactive source could be given as, 

$\frac{dN}{dt}=8.0mCi$

$\Rightarrow \frac{dN}{dt}=8\times {{10}^{-3}}\times 3.7\times {{10}^{10}}=29.6\times {{10}^{7}}decay/s$

Where, N is the required number of atoms. 

Half life of ${}_{27}^{60}Co$, ${{T}_{\frac{1}{2}}}=5.3years$

$\Rightarrow {{T}_{\frac{1}{2}}}=5.3\times 365\times 24\times 60\times 60=1.67\times {{10}^{8}}s$

For decay constant $\lambda $, we could give the rate of decay as, 

$\frac{dN}{dt}=\lambda N$

Where, $\lambda =\frac{0.693}{{{T}_{\frac{1}{2}}}}=\frac{0.693}{1.67\times {{10}^{8}}}{{s}^{-1}}$

$\Rightarrow N=\frac{1}{\lambda }\frac{dN}{dt}=\frac{29.6\times {{10}^{7}}}{\left( \frac{0.693}{1.67\times {{10}^{8}}} \right)}=7.133\times {{10}^{16}}atoms$

Now for ${}_{27}C{{o}^{60}}$, Mass of Avogadro number of atoms $=60g$

Then, mass of $7.133\times {{10}^{16}}atoms=\frac{60\times 7.133\times {{10}^{16}}}{6.023\times {{10}^{23}}}=7.106\times {{10}^{-6}}g$

Therefore, the amount of ${}_{27}C{{o}^{60}}$that is required for the purpose is $7.106\times {{10}^{-6}}g$. 

Ans: We know that, 

Half life of ${}_{38}^{90}Sr$, ${{t}_{\frac{1}{2}}}=28years=28\times 365\times 24\times 3600=8.83\times {{10}^{8}}s$

Mass of the isotope, $m=15mg$

90g of ${}_{38}^{90}Sr$ atom contains Avogadro number of atoms. So, 15mg of ${}_{38}^{90}Sr$ contains, 

$\frac{6.023\times {{10}^{23}}\times 15\times {{10}^{-3}}}{90}=1.0038\times {{10}^{20}}\text{number of atoms}$

Rate of disintegration would be, $\frac{dN}{dt}=\lambda N$

Where, $\lambda $is the decay constant given by, $\lambda =\frac{0.693}{8.83\times {{10}^{8}}}{{s}^{-1}}$

$\therefore \frac{dN}{dt}=\frac{0.693\times 1.0038\times {{10}^{20}}}{8.83\times {{10}^{8}}}=7.878\times {{10}^{10}}atoms/s$

Therefore, we found the disintegration rate of 15mg of given isotope to be $7.878\times {{10}^{10}}atoms/s$.