NCERT Solutions for Class 12 Physics Chapter 13 Nuclei

1.1 Isotopes

The atoms of an element, which have the same atomic number but different mass numbers, are called isotopes.

(i) 8O16, 8O17, 8O18

(ii) 17Cl35, 17Cl37

(iii) 82Pb206, 82Pb207, 82Pb208.

1.2 Isotones

The atoms whose nuclei have same number of neutrons are called isotones. 6C14, 7N15

1.3 Isobars

The atoms, which have same mass number but different atomic numbers, are called isobars.

(i) 1H3 and 1H3

(ii) 2Li7 and 4Be7

(iii) 28Ar40 and 29Ca40

(iv) 32Ge76 and 34Se76

1.4 Atomic Mass Unit

The atomic mass unit (a.m.u) is a very small unit of mass and it is found to be very convenient in nuclear physics.

Atomic mass unit is defined as 1/12th of the mass of one 6C12 atom.

According to Avogadro’s hypothesis, number of atoms in 12 g of 6C12 is equal to Avogadro number i.e. 6.023 × 1023.

Therefore, the mass of one carbon atom (6C12) is i.e. 1.992678 × 10–26 kg.

Therefore, 1 a.m.u. 

or 1 a.m.u. = 1.660565 × 10-27 kg

1.5 Energy Equivalent of Atomic Mass Unit

According to Einstein’s mass-energy equivalence relation, the energy equivalent of mass m is given by E = mc2

Where c is speed of light.

Suppose that m = 1 a.m.u = 1.660565 × 10-27 kg

Since, c = 2.998 × 108 ms-1, the energy equivalent of 1 a.m.u is given by 

1 a.m.u = (1.660565 × 10-27 kg) × (2.998 × 108 ms1)2
= 1.4925 × 10-10 J

Since, 1 MeV = 1.602 × 10-13 J, we have 

or 1 a.m.u = 931.5 MeV

1.6 Nuclear Size

The volume of the nucleus is directly proportional to the number of nucleons (mass number) constituting the nucleus. If R is the radius of the nucleus having mass number A, then

$\frac{4}{3}\pi R^3\alpha A$

or R α A1/3 or R = R0 A1/3       R0 = 1.2 × 10-15m

1.7 Nuclear Density

Mass of the nucleus of the atom of mass number A = A a.m.u = A × 1.660565 × 10-27 kg. If R is radius of the nucleus, then

Volume of nucleus = $\frac{4}{3}\pi R^3=\frac{4}{3}(R_0^3{A}^{1/3}{)}^3=\frac{4}{3}\pi R_0^{3}A$

Taking R0 = 1.2 × 10-15 m, we have

Density of the nucleus, $\rho =\frac{\text{mass of nucleus}}{\text{volume of nucleus}}$ 

$=\frac{A\times 1.66065\times {10}^{-27}}{\frac{4}{3}\pi (1.2\times {10}^{-15}{)}^3\times A}=2.26\times {10}^{17}\text{kg}{m}^{-3}$

(independent of A)

NOTE:

1.8 Mass Defect

The difference between the sum of the masses of the nucleons constituting a nucleus and the rest mass of the nucleus is known as mass defect. It is denoted by Δm.

Let us calculate the mass defect in case of the nucleus of an atom zXA. The nucleus of the atom contains Z protons and (A−Z) neutrons. Therefore, if mN (zXA) is mass of the nucleus of the atom ZXA, then the mass defect is given by

Δm = [Zmp + (A−Z)mn −mN (zXA)]

The binding energy of a nucleus may be defined as the energy equivalent to the mass defect of the nucleus. It may be measured as the work required to be done to separate the nucleon an infinite distance apart, so that they no longer interact with each other.

If Δm is mass defect of a nucleus, then according to Einstein’s mass-energy relation, binding energy of the nucleus = Δm c2 (in joule).

Here, mass defect Δm has to be measured in kilogram.  In case, mass defect is measured in a.m.u., then

Binding energy of the nucleus = Δm × 931.5 (in MeV)

Binding energy = [Zmp + (A−Z)mn− mN (ZXA)] × 931.5 MeV

1.9 Binding Energy Per Nucleon

The binding energy per nucleon is the average energy required to extract one nucleon from the nucleus.

Thus, binding energy per nucleon $={\displaystyle \frac{bindingenergy}{A}}$

Fig. 10.9

1.10 Packing Fraction

Packing fraction = (mass defect)/A.

1.11 Nuclear Fission

The process of splitting of a heavy nucleus into two nuclei of nearly comparable masses with liberation of energy is called nuclear fission.

92U235 + 0n1 → [92U236] → 56Ba141+ 36Kr92 + 30n1 + Q

Neutron reproduction factor is defined as the ratio of the rate of production of neutrons to the rate of loss of neutrons. Thus, $\text{k}=\frac{\text{rate of production of neutrons}}{\text{rate of loss of neutrons}}$

A fission reaction will be steady, in case k = 1. In case k > 1, the fission reaction will accelerate and it will retard, in case  k < 1.

1.12 Nuclear Fusion

When two light nuclei fuse to form a larger nucleus, energy is released, this process is called nuclear fusion. Some examples of such energy liberating fusion reactions are:

${}_1^1\text{H}{+}_1^1\text{H}{\to }_1^2\text{H}+{\text{e}}^{+}+\text{v}+0.42\text{MeV}$

${}_1^2\text{H}{+}_1^2\text{H}{\to }_2^3\text{He}+\text{n}+3.27\text{MeV}$

${}_1^2\text{H}{+}_1^2\text{H}{\to }_1^3\text{H}+{}_1^1\text{H}+4.03\text{MeV}$

In the first reaction, two protons combine to form a deuteron and a position with a release of 0.42 MeV energy. In second reaction, two deuterons combine to form the light isotope of helium. In third reaction, two deuterons combine to form a triton and a proton. For fusion to take place, the two nuclei must come close enough so that attractive short – range nuclear force is able to affect them. However, since they are both positively charged particles, they experience coulomb repulsion. They, therefore, must have enough energy to overcome this coulomb barrier. The height of the barrier depends on the charges and radii of the two interacting nuclei. It can be shown, for example, that the barrier height for two protons is ~ 400 keV, and is higher for nuclei with higher charges. We can estimate the temperature at which two protons in a proton gas would (averagely) have enough to overcome the coulomb barrier :

$\left(\frac{3}{2}\right)\text{kT}=\text{K}$ 400 keV, which gives T ~ 3 × 109 K. 

When fusion is achieved by raising the temperature of the system so that particles have enough kinetic energy to overcome the coulomb repulsive behaviour, it is called thermonuclear fusion.

2.1 Laws of Radioactivity Decay

Rutherford and Soddy studied the phenomenon of radioactivity in details and formulated the following laws, known as the laws of radioactive decay:

• Radioactivity is a spontaneous phenomenon and one cannot predict, when a particular atom in a given radioactive sample will undergo disintegration.

• When a radioactive atom disintegrates, either an α-particle (nucleus of helium) or a β-particle (electron) is emitted.

• The emission of an α-particle by a radioactive atom result in a daughter atom, whose atomic number is 2 units less and mass number is 4 units less than that of the parent atom.

$z^{X^A}\stackrel{\alpha -\text{decay}}{\to }z-2^{Y^{A-4}}$

• The emission of a β-particle by a radioactive atom result in a daughter atom, whose atomic number is 1 unit more but mass number is same as that of the parent atom.

$z^{X^A}\stackrel{\beta -\text{decay}}{\to }z+1^{Y^A}$

• The number of atoms disintegrating per second of a radioactive sample at any time is directly proportional to the number of atoms present at that time.  The rate of disintegration of the sample cannot be altered by changing the external factors, such as pressure, temperature etc.  It is known as radioactive decay law.

According to radioactive decay law, the rate of disintegration at any time t is directly proportional to the number of atoms present at time t i.e.

$\frac{\text{dN}}{\text{dt}}\propto \text{N}\text{or}\frac{\text{dN}}{\text{dt}}=-\lambda \text{N}.$

Where the constant of proportionally λ is called decay constant of the radioactive sample.  It is also known as disintegration constant or transformation constant.  Its value depends upon the nature of the radioactive sample.  Further, the negative sign indicates that the number of the atoms of the sample decreases with the passage of time.

From equation, we have $\frac{\text{dN}}{\text{dt}}=-\lambda \text{dt}.$

Integrating ${\int }_{N_0}^N\frac{dN}{N}={\int }_0^t\lambda dt$

Or ${\log }_e\frac{N}{N_0}=-\lambda t$

Or $\frac{N}{N_0}=e^{-\lambda t}$

Or N = N0e−λt

2.3 Half Life

Consider that a radioactive sample contains N0 atoms at time t = 0.  Then, the number of atoms left behind after time t is given by N = N0 e−λ t

From the definition of half-life, it follows that when t = t1/2, $N=\frac{N_0}{2}.$

Setting the above condition in equation, we have $\frac{N_0}{2}=N_0{e}^{-\lambda t_{1/2}}$

Or $e^{-\lambda t_{1/2}}=1/2$ Or e^{\lambda t_{1/2}} = 2

Or  λT = loge 2 = 2.303 log10 2 = 2.303 × 0.3010 = 0.693

Or $t_{1/2}=\frac{0.693}{\lambda }$

Thus, half life of a radioactive substance is inversely proportional to its decay constant and is a characteristic property of its nucleus. It cannot be altered by any known method.

2.4 Mean life or Average Life

The average life of a radioactive substance is defined as the average time for which the nuclei of the atoms of the radioactive substance exist. It is defined by tavg.

$t_{\text{avg}}=\frac{1}{\lambda }$

Mastering Class 12 Physics Chapter 13: Nuclei - MCQs, Question and Answers, and Tips for Success

1. Questions and Answers

(a) Two stable isotopes of lithium ${}_3^{6}Li$and ${}_3^{7}Li$ have respective abundances of $7.5$ and $92.5$These isotopes have masses $6.01512u$and$7.01600u$, respectively. Find the atomic mass of lithium.

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 $6.940934u$. 

(b) Boron has Two Stable Isotopes ${}_5^{10}B$ and${}_5^{11}B$. Their respective masses are $10.01294u$and$11.00931u$, and the atomic mass of boron is $10.811u$. Find the abundances of ${}_5^{10}B$ and ${}_5^{11}B$ .

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  ${}_5^{11}B$, $n_2=(100-x)$

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,

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

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

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

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. 

2. The three stable isotopes of neon: ${}_{10}^{20}Ne,{}_{10}^{21}Ne\text{ and }{}_{10}^{22}Ne$ have respective abundances of $$90.51$$The atomic masses of the three isotopes are$$\mathbf{19}.\mathbf{99}\mathbf{u},\mathbf{20}.\mathbf{99}\mathbf{u}\mathbf{and}\mathbf{21}.\mathbf{99}\mathbf{u}$$, respectively. Obtain the average atomic mass of neon.

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 $20.177u.$

3. Obtain the binding energy (in MeV) of a nitrogen nucleus $\left({}_7^{14}N\right)$, given $m\left({}_7^{14}N\right)=14.00307u$

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, $\mathrm{\Delta }m=7m_H+7m_n-m$

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, 

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

$\Rightarrow \mathrm{\Delta }m=7.054775+7.06055-14.00307$

$\therefore \mathrm{\Delta }m=0.11236u$

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

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

Now, we could give the binding energy as, 

$E_b=\mathrm{\Delta }m{c}^2$

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 $104.66334MeV$. 

4. Obtain the binding energy of the nuclei ${}_{26}^{56}Fe$ and ${}_{83}^{209}Bi$ in units of MeV from the following data: $m\left({}_{26}^{56}Fe\right)=55.934939u$, $m\left({}_{83}^{209}Bi\right)=208.980388u$

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, $\mathrm{\Delta }m=26\times m_H+30\times m_n-m_1$

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, 

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

$\Rightarrow \mathrm{\Delta }m=26.20345+30.25995-55.934939$

$\therefore \mathrm{\Delta }m=0.528461u$

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

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

The binding energy of this nucleus could be given as,

$E_{b1}=\mathrm{\Delta }m{c}^2$

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, $m_n=1.008665u$

$\mathrm{\Delta }{m}^{\prime }=83\times 1.007825+126\times 1.008665-208.980388$

$\Rightarrow \mathrm{\Delta }{m}^{\prime }=83.649475+127.091790-208.980388$

$\therefore \mathrm{\Delta }{m}^{\prime }=1.760877u$

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

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

$E_{b2}=\mathrm{\Delta }{m}^{\prime }{c}^2=1.760877\times 931.5\left(\frac{MeV}{c^2}\right)\times c^2$

$\therefore E_{b2}=1640.26MeV$

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 $7.848MeV$. 

5. A given coin has a mass of 3.0 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 ${}_{29}^{63}Cu$ atoms (of mass$$\mathbf{62}.\mathbf{92960}\mathbf{u}$$).

Ans: We are given:

Mass of a copper coin, $m^{\prime }=3g$

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

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

Where,$$N_A=Avogadronumber=6.023\times {10}^{23}atoms/g$$

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 $$(63-29)=34$$neutrons

Mass defect of this nucleus would be, $\mathrm{\Delta }{m}^{\prime }=29\times m_H+34\times m_n-m$

Where,

Mass of a proton, $m_H=1.007825u$

Mass of a neutron, $m_n=1.008665u$$\mathrm{\Delta }{m}^{\prime }=29\times 1.007825+34\times 1.008665-62.9296=0.591935u$

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

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

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

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

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

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

$\therefore E_b=1.581\times {10}^{25}MeV$

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.

6. Write the Nuclear Reactions For:

(a) $\alpha -\text{decay of }{}_{88}^{226}Ra$

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, 

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

(b) $\alpha -\text{decay of }{}_{94}^{242}Pu$

Ans: ${}_{94}^{242}Pu\to {}_{92}^{238}U+{}_2^{4}He$

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

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

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

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

3. ${\beta }^{+}-\text{decay of }{}_6^{11}C$

Ans: ${}_6^{11}C\to {}_5^{11}B+e^{+}+\nu$

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

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

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

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

7. A radioactive isotope has a half-life of T years. How long will it take the activity to reduce to:

1. 3.125% of Its Original Value? 

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,

$\frac{N}{N_0}=3.125$

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. 

2. 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:

$\frac{N}{N_0}=1$

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. 

8. The normal activity of living carbon-containing matter is found to be about 15 decays per minute for every gram of carbon. This activity arises from the small proportion of radioactive ${}_6^{14}C$ present with the stable carbon isotope ${}_6^{12}C$ . When the organism is dead, its interaction with the atmosphere (which maintains the above equilibrium activity) ceases and its activity begins to drop. From the known half-life (5730 years) of ${}_6^{14}C$ , and the measured activity, the age of the specimen can be approximately estimated. This is the principle of ${}_6^{14}C$ dating used in archaeology. Suppose a specimen from Mohenjodaro gives an activity of 9 decays per minute per gram of carbon. Estimate the approximate age of the Indus-Valley civilisation.

Ans: We are given that: 

Decay rate of living carbon-containing matter, $$R=15decay/min$$

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

Half life of ${}_6^{14}C,{\text{T}}_{\frac{1}{2}}=5730\text{ years}$

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

$$R^{\prime }=9decays/min$$

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:

$\frac{N^{\prime }}{N}=\frac{R^{\prime }}{R}=e^{-\lambda t}$

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

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

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

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.

9. Obtain the Amount of ${}_{27}^{60}Co$ necessary to provide a radioactive source of $8.0\text{ mCi}$strength. The half-life of ${}_{27}^{60}Co$ is 5.3 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 $7.106\times {10}^{-6}g$. 

10. The half life of ${}_{38}^{90}Sr$ is 28years. What is the disintegration rate of 15mg of this isotope?

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 $7.878\times {10}^{10}atoms/s$.

11. Obtain approximately the ratio of the nuclear radii of the gold isotope ${}_{79}^{197}Au$ and the silver isotope ${}_{47}^{107}Ag$.

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.

12. Find the Q-value and the kinetic energy of the emitted α-particle in the α-decay of: [Given:$m\left({}_{88}^{226}Ra\right)=226.02540u,\text{ m}\left({}_{89}^{222}Rn\right)=222.01750u,$ $\text{m}\left({}_{86}^{220}Rn\right)=220.01137u,\text{ m}\left({}_{84}^{216}Po\right)=216.00189u$ ]

1. ${}_{88}^{226}Ra$

Ans: We know that, 

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

${}_{88}^{226}Ra\to {}_{86}^{222}Ra+{}_2^{4}He$

$$Q-valueofemitted\alpha -particle=(Sumofinitialmass-Sumoffinalmass)c^2$$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, 

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

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

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 $4.85MeV$.

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

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, 

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

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 $6.29MeV$. 

13. The Radionuclide ${}_6^{11}C$ decays according to,

${}_6^{11}C\to {}_5^{11}B+e^{+}+\nu \text{ ; }{\text{T}}_{\frac{1}{2}}=20.3min$

The maximum energy of the emitted positron is $$\mathbf{0}.\mathbf{960}\mathbf{MeV}$$.

Given the mass values: $m\left({}_6^{11}C\right)=11.011434u\text{ and m}\left({}_6^{11}B\right)=11.009305u$

Calculate Q and compare it with the maximum energy of the positron emitted. 

Ans: The given nuclear reaction is, 

${}_6^{11}C\to {}_5^{11}B+e^{+}+\nu$

Half life of ${}_6^{11}C$ nuclei, $T_{\frac{1}{2}}=20.3min$

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 $$\left(\mathrm{\Delta }Q\right)$$ of the nuclear masses of the ${}_6^{11}C$

$\mathrm{\Delta }Q=[m^{\prime }\left({}_6{C}^{11}\right)-[m^{\prime }\left({}_5^{11}B\right)+m_e]]c^2$…………………….. (1)

Where,$m_e=$Mass of an electron or positron = $$0.000548u$$

$c=$ Speed of light

$m^{\prime }=$Respective nuclear masses

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

Hence, equation (1) would now reduce to,

$\mathrm{\Delta }Q=[m\left({}_6{C}^{11}\right)-m\left({}_5^{11}B\right)-2m_e]c^2$

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, 

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

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

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

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

14. The nucleus  ${}_{10}^{23}Ne$ decays by ${\beta }^{-}$emission. Write down the $\beta$decay equation and determine the maximum kinetic energy of the electrons emitted. Given that:

$m\left({}_{10}^{23}Ne\right)=22.994466u$

$m\left({}_{11}^{23}Na\right)=22.989770u$

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, 

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

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:

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

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.

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

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 $e^{-}\text{ and }\overline{\nu }$. 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.,$$4.374MeV$$.

15. The Q-value of a nuclear reaction $A+b\to C+d$ is defined by $Q=[m_A+m_b-m_C-m_d]c^2$ 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.

Atomic masses are given to be: $m\left({}_1^{2}H\right)=2.014102u,\text{ m}\left({}_1^{3}H\right)=3.016049u,$

$m\left({}_6^{12}C\right)=12.000000u,\text{ m}\left({}_{10}^{20}Ne\right)=19.992439u$

1. ${}_1^{1}H+{}_1^{3}H\to {}_1^{2}H+{}_1^{2}H$

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:

$Q=[m\left({}_1^{1}H\right)+m\left({}_1^{3}H\right)-2m\left({}_1^{2}H\right)]c^2$

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

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. 

2. ${}_6^{12}C+{}_6^{12}C\to {}_{10}^{20}Ne+{}_2^{4}He$

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, 

$Q=[2m\left({}_6^{12}C\right)-m\left({}_{10}^{20}Ne\right)-m\left({}_2^{4}He\right)]c^2$

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