Nuclei: Overview, Questions, Preparation

Q.13.1 (a) Two stable isotopes of lithium  ${}_3{}^6\mathit{L}\mathit{i}$ and  ${}_3{}^7\mathit{L}\mathit{i}$ have respective abundances of 7.5% and 92.5%. These isotopes have masses 6.01512 u and 7.01600 u, respectively. Find the atomic mass of lithium.

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

Ans.13.1 Mass of  ${}_3{}^{6}Li$ lithium isotope,  $m_1$ = 6.01512 u

Mass of  ${}_3{}^{7}Li$ lithium isotope,  $m_2$ = 7.01600 u

Abundance of  ${}_3{}^{6}Li$ ,  ${\eta }_1$ = 7.5%

Abundance of  ${}_3{}^{7}Li$ ,  ${\eta }_2$ = 92.5%

The atomic mass of lithium atom is given as:

m =  $\frac{m_1{\eta }_1+m_{2{\eta }_2}}{{\eta }_1+{\eta }_2}$ =  $\frac{6.01512\mathrm{}\times 7.5+\mathrm{}7.01600\mathrm{}\times 92.5}{7.5+92.5}$ = 6.940934 u

Mass of  ${}_5{}^{10}B$ Boron isotope,  $m_1$ = 10.01294 u

Mass of  ${}_5{}^{11}B$ Boron isotope,  $m_2$ = 11.00931 u

Let the abundance of  ${}_5{}^{10}B$ be x % and that of  ${}_5{}^{11}B$ be (100-x) %

The atomic mass of Boron atom is given as :

10.8111 =  $\frac{10.01294x+11.00931(100-x)}{x+(100-x)}$

1081.11 = 1100.931 - 0.99637x

x = 19.89 %

Hence the abundance of  ${}_5{}^{10}B$ is 19.89 % and that of  ${}_5{}^{11}B$ is (100-19.89) = 80.11 %

 

 

Q.13.2 The three stable isotopes of neon:  ${}_{10}{}^{20}\mathit{N}\mathit{e},$  ${}_{10}{}^{21}\mathit{N}\mathit{e}$ and  ${}_{10}{}^{22}\mathit{N}\mathit{e}$ have respective abundances of 90.51%, 0.27% and 9.22%. The atomic masses of the three isotopes are 19.99 u, 20.99 u and 21.99 u, respectively. Obtain the average atomic mass of neon.

Ans.13.2 Atomic mass of  ${}_{10}{}^{20}Ne$ neon isotope,  $m_1$ = 19.99 u ad the abundance  ${\eta }_1$ = 90.51 %

Atomic mass of  ${}_{10}{}^{21}Ne$ neon isotope,  $m_2$ = 20.99 u ad the abundance  ${\eta }_2$ = 0.27 %

Atomic mass of  ${}_{10}{}^{22}Ne$ neon isotope,  $m_3$ = 21.99 u ad the abundance  ${\eta }_3$ = 9.22 %

The average atomic mass of neon is given as:

m =  $\frac{m_1{\eta }_1+m_{2{\eta }_2}+m_3{\eta }_3}{{\eta }_1+{\eta }_2+{\eta }_3}$ =  $\frac{19.99\times 90.51+20.99\times 0.27+21.99\times 9.22}{90.51+0.27+9.22}$ =  $\frac{2017.71}{100}$ = 20.1771 u

 

Q.13.3 Obtain the binding energy (in MeV) of a nitrogen nucleus (  ${}_7{}^{14}\mathit{N})$ , given

m(  ${}_7{}^{14}\mathit{N})$ =14.00307 u

Ans.13.3 Atomic mass of  ${}_7{}^{14}N$ nitrogen , m = 14.00307 u

A nucleus of  ${}_7{}^{14}N$ nitrogen contains 7 protons and 7 neutrons.

Hence, the mass defect of this nucleus, Δm = 7  $m_p$ + 7  $m_n$ - m, where

Mass of a proton,  $m_p$ = 1.007825 u

Mass of a neutron,  $m_n$ = 1.008665 u

Therefore, Δm = 7  $\times$ 1.007825+ 7  $\times$ 1.008665 – 14.00307 = 0.11236 u

But 1 u = 931.5 MeV/  $c^2$

Δm = 104.66334 MeV/  $c^2$

The binding energy of the nucleus,  $E_b$ = Δm  $c^2$ , where c = speed of light

$E_b=$ (104.66334/  $c^2$ )  $\times$  $c^2$ = 104.66334 MeV

 

Q.13.4 Obtain the binding energy of the nuclei  ${}_{26}{}^{56}\mathit{F}\mathit{e}$ and  ${}_{83}{}^{209}\mathit{B}\mathit{i}$ in units of MeV from the following data:

m (  ${}_{26}{}^{56}\mathit{F}\mathit{e})$ = 55.934939 u m (  ${}_{83}{}^{209}\mathit{B}\mathit{i}$ ) = 208.980388 u

Ans.13.4 Atomic mass of  ${}_{26}{}^{56}Fe$ ,  $m_1$ = 55.934939 u

${}_{26}{}^{56}Fe$ has 26 protons and (56-26) 30 neutrons

Hence the mass defect of the nucleus Δm = 26  $\times m_p$ + 30  $\times m_n$ -  $m_1$

Mass of a proton,  $m_p$ = 1.007825 u

Mass of a neutron,  $m_n$ = 1.008665 u

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

Δm = 0.528461 u

But 1 u = 931.5 MeV/  $c^2$

Δm = 492.2614215 MeV/  $c^2$

The binding energy of the nucleus,  $E_b$ = Δm  $c^2$ , where c = speed of light

$E_b=$ (492.2614215 /  $c^2$ )  $\times$  $c^2$ = 492.2614215 MeV

Average binding energy per nucleon =  $\frac{E_b}{56}$ = 8.79 MeV

Atomic mass of  ${}_{83}{}^{209}Bi$ ,  $m_2$ = 208.980388 u

${}_{83}{}^{209}Bi$ have 83 protons and (209-83) 126 neutrons

Hence the mass defect of the nucleus Δm = 83  $\times m_p$ + 126  $\times m_n$ -  $m_2$

Mass of a proton,  $m_p$ = 1.007825 u

Mass of a neutron,  $m_n$ = 1.008665 u

Δm = 83  $\times 1.007825\mathrm{}$ + 126  $\times 1.008665\mathrm{}$ - 208.980388

Δm = 1.760877 u

But 1 u = 931.5 MeV/  $c^2$

Δm = 1640.256926 MeV/  $c^2$

The binding energy of the nucleus,  $E_b$ = Δm  $c^2$ , where c = speed of light

$E_b=$ (1640.256926 /  $c^2$ )  $\times$  $c^2$ = 1640.256926 MeV

Average binding energy per nucleon =  $\frac{E_b}{209}$ = 7.848 MeV

 

Q.13.5 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  ${}_{29}{}^{63}\mathit{C}\mathit{u}$ atoms (of mass 62.92960 u).

Ans.13.5 Mass of the copper coin, m’ = 3.0 g

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

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

$N_A$ = Avogadro’s number = 6.023  $\times {10}^{23}$ atoms / g

Mass number = 63 g

Therefore, N =  $\frac{6.023\mathrm{}\times {10}^{23}\times 3}{63}$ = 2.868  $\times {10}^{22}$ atoms

${}_{29}{}^{63}Cu$ has 29 protons and (63 – 29) 34 neutrons

Hence the mass defect of the nucleus Δm = 29  $\times m_p$ + 34  $\times m_n$ -  $m$

Mass of a proton,  $m_p$ = 1.007825 u

Mass of a neutron,  $m_n$ = 1.008665 u

Δm = 29  $\times 1.007825\mathrm{}$ + 34  $\times 1.008665\mathrm{}$ - 62.92960

Δm = 0.591935 u

Mass defect of all the atoms present in the coin, Δm = 0.591935  $\times N$

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

But 1 u = 931.5 MeV/  $c^2$

Δm = 1.581  $\times {10}^{25}$ MeV/  $c^2$

The binding energy of the nucleus,  $E_b$ = Δm  $c^2$ , where c = speed of light

$E_b=$ (1.581  $\times {10}^{25}$ /  $c^2$ )  $\times$  $c^2$ = 1.581  $\times {10}^{25}$ MeV

1 MeV = 1.6  $\times {10}^{-13}$ J

Hence,  $E_b=$ 2.530  $\times {10}^{12}$ J

This much of energy required to separate all the neutrons and protons from the given coin.

 

Q.13.6 Write nuclear reaction equations for

(i)  $\mathit{\alpha }$ -decay of  ${}_{88}{}^{226}\mathit{R}\mathit{a}$

(ii)  $\mathit{\alpha }$ -decay of  ${}_{94}{}^{242}\mathit{P}\mathit{u}$

(iii)  $\mathit{\beta }$ –-decay of  ${}_{15}{}^{32}\mathit{P}$

(iv)  $\mathit{\beta }$ –-decay of  ${}_{83}{}^{210}\mathit{B}\mathit{i}$

(v)  $\mathit{\beta }$ +decay of  ${}_6{}^{11}\mathit{C}$

(vi)  $\mathit{\beta }$ –-decay of  ${}_{43}{}^{97}\mathit{T}\mathit{c}$

(vii) Electron capture of  ${}_{54}{}^{120}\mathit{X}\mathit{e}$

Ans.13.6 In  $\alpha$ - decay, there is a loss of 2 protons and 4 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.

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

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

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

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

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

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

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

 

Q.13.7 A radioactive isotope has a half-life of T years. How long will it take the activity to reduce to a) 3.125%, b) 1% of its original value?

Ans.13.7 Half life of the radioactive isotope = T years

Original amount of the radioactive isotope =  $N_o$

After decay, the amount of radioactive isotope = N

It is given that only 3.125% of  $N_o$ remains after decay. Hence, we can write,

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

But  $\frac{N}{N_o}$ =  $e^{-\lambda t}$ , where  $\lambda$ = decay constant, t = time

Therefore,

$e^{-\lambda t}=\frac{1}{32}$

By taking log on both sides

${\log }_e{e}^{-\lambda t}$ =  ${\log }_e\frac{1}{32}$

$-\lambda t=$  ${\log }_{e}1$ -  ${\log }_{e}32$

$-\lambda t$ = 0 – 3.465

$t$ =  $\frac{3.465}{\lambda }$

Since  $\lambda$ =  $\frac{0.693}{T}$

t =  $\frac{3.465}{\frac{0.693}{T}}$ = 5T years

Hence, all the isotopes will take about 5T years to reduce 3.125% of its original value.

After decay, the amount of radioactive isotope = N

It is given that only 1.0% of  $N_o$ remains after decay. Hence, we can write,

$\frac{N}{N_o}$ = 1% =  $\frac{1}{100}$

But  $\frac{N}{N_o}$ =  $e^{-\lambda t}$ , where  $\lambda$ = decay constant, t = time

Therefore,

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

By taking log on both sides

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

$-\lambda t=$  ${\log }_{e}1$ -  ${\log }_{e}100$

$-\lambda t$ = 0 – 4.605

$t$ =  $\frac{4.605}{\lambda }$

Since  $\lambda$ =  $\frac{0.693}{T}$

t =  $\frac{4.605}{\frac{0.693}{T}}$ = 6.645T years

Hence, all the isotopes will take about 6.645T years to reduce 1.0% of its original value.

 

Q.13.8 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  ${}_6{}^{14}\mathit{C}$ present with the stable carbon isotope  ${}_6{}^{12}\mathit{C}$ . 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  ${}_6{}^{14}\mathit{C}$ , and the measured activity, the age of the specimen can be approximately estimated. This is the principle of  ${}_6{}^{14}\mathit{C}$ dating used in archaeology. Suppose a specimen from Mohenjo-Daro gives an activity of 9 decays per minute per gram of carbon. Estimate the approximate age of the Indus-Valley civilization.

Ans.13.8 Decay rate of living carbon-containing matter, R = 15 decay / min

Half life of  ${}_6{}^{14}C$ ,  $T_{1/2}$ = 5730 years

Decay rate of the specimen obtained from the Mohenjo-Daro site, R’ = 9 decays/min

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

Let N’ be the number of radioactive atoms present in the specimen during the Mohenjo-Daro period.

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

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

$e^{-\lambda t}=$  $\frac{R\text{'}}{R}$ =  $\frac{9}{15}$ =  $\frac{3}{5}$

By taking log (ln) on both sides,

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

t =  $\frac{0.5108}{\lambda }$

Since  $\lambda$ =  $\frac{0.693}{T_{1/2}}$ =  $\frac{0.693}{5730}$

t =  $\frac{5730\times 0.5108}{0.693}$ = 4223.5 years

Hence, the approximate age of the Indus-valley is 4223.5 years.

 

Q.13.9 Obtain the amount of  ${}_{27}{}^{60}\mathit{C}\mathit{o}$ necessary to provide a radioactive source of 8.0 mCi strength. The half-life of  ${}_{27}{}^{60}\mathit{C}\mathit{o}$ is 5.3 years.

Ans.13.9 The strength of the radioactive source is given as:

$\frac{dN}{dt}$ = 8.0 mCi = 8  $\times {10}^{-3}$  $\times 3.7\times {10}^{10}$ decay/s = 296  $\times {10}^6$ decay/s, where

N = Required number of atoms

Given, half life of  ${}_{27}{}^{60}Co$ ,  $T_{1/2}$ = 5.3 years = 5.3  $\times 365\times 24\times 60\times 60$ secs = 167  $\times {10}^6$ s

For decay constant  $\lambda$ , we have rate of decay as:

$\frac{dN}{dt}$ =  $\lambda N$ or

N =  $\frac{1}{\lambda }\frac{dN}{dt}$ , where  $\lambda$ =  $\frac{0.693}{T_{1/2}}$ =  $\frac{0.693}{167\mathrm{}\times {10}^6}$ /s = 4.1497  $\times {10}^{-9}$  $s^{-1}$

N =  $\frac{296\mathrm{}\times {10}^6\mathrm{}}{4.1497\times {10}^{-9}}$ = 7.133  $\times {10}^{16}$ atoms

For  ${}_{27}{}^{60}Co$ , mass of 6.023  $\times {10}^{23}$ atoms = 60 gms

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

 

Q.13.10 The half-life of  ${}_{38}{}^{90}\mathit{S}\mathit{r}$ is 28 years. What is the disintegration rate of 15 mg of this isotope?

Ans.13.10 Half life of  ${}_{38}{}^{90}Sr$ ,  $T_{1/2}$ = 28 years = 28  $\times 365\times 24\times 60\times 60$ secs = 0.883  $\times {10}^9$ s

Mass of the isotope, m = 15 mg = 15  $\times {10}^{-3}$ gms

90 g of  ${}_{38}{}^{90}Sr$ contains 6.023  $\times {10}^{23}$ atoms

No. of atoms in 15 mg of  ${}_{38}{}^{90}Sr$ contains =  $\frac{6.023\times {10}^{23}}{90}\times$ 15  $\times {10}^{-3}$ = 1.0038  $\times {10}^{20}$

Rate of disintegration  $\frac{dN}{dt}$ =  $\lambda N$ , where  $\lambda$ =  $\frac{0.693}{T_{1/2}}$ =  $\frac{0.693}{0.883\times {10}^9}$ /s = 7.848  $\times {10}^{-10}$  $s^{-1}$

$\frac{dN}{dt}$ = 7.848  $\times {10}^{-10}\times$ 1.0038  $\times {10}^{20}$ = 7.878  $\times {10}^{10}$ atoms / second.

 

Q.13.11 Obtain approximately the ratio of the nuclear radii of the gold isotope  ${}_{47}{}^{197}\mathit{A}\mathit{u}$

and the silver isotope  ${}_{47}{}^{107}\mathit{A}\mathit{g}$ .

Ans.13.11 Nuclear radius of the gold isotope,  ${}_{47}{}^{197}Au$ =  $R_{Au}$

Nuclear radius of silver isotope,  ${}_{47}{}^{107}Ag$ =  $R_{Ag}$

Mass number of gold,  $A_{Au}$ = 197

Mass number of silver,  $A_{Ag}$ = 107

The ratio of the radii of the two nuclei is related with their mass numbers as :

$\frac{R_{Au}}{R_{Ag}}$ =  ${\left[\frac{A_{Au}}{A_{Ag}}\right]}^{1/3}$ =  ${\left[\frac{197}{107}\right]}^{1/3}$ =1.2256

Hence, the ratio of the nuclear radii of the gold and silver isotope is 1.23

 

Q.13.12 Find the Q-value and the kinetic energy of the emitted  $\mathit{\alpha }$ -particle in the  $\mathit{\alpha }$ -decay of (a)  ${}_{88}{}^{226}\mathit{R}\mathit{a}$ and (b)  ${}_{86}{}^{220}\mathit{R}\mathit{n}$ . Given m (  ${}_{88}{}^{226}\mathit{R}\mathit{a})$ = 226.02540 u, m (  ${}_{86}{}^{222}\mathit{R}\mathit{n})$ = 222.01750 u,

m (  ${}_{86}{}^{222}\mathit{R}\mathit{n})\mathit{}$ = 220.01137 u, m (  ${}_{84}{}^{216}\mathit{P}\mathit{o})\mathit{}$ = 216.00189 u.

Ans.13.12 $\alpha -$ particle decay of  ${}_{88}{}^{226}Ra$ emits a helium nucleus. As a result, its mass number reduces to (226-4) 222 and its atomic number reduces to (88-2) 86.

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

Q value of emitted  $\alpha -$ particle = (Sum of initial mass – Sum of final mass)  $\times c^2$ , where

c = Speed of light.

It is given that

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

m (  ${}_{86}{}^{222}Rn)$ = 222.01750 u

m (  ${}_2{}^{4}He)$ = 4.002603 u

Q value = [(226.02540) – (222.01750 + 4.002603)]  $c^2$

= 5.297  $\times {10}^{-3}{uc}^2$

But 1 u = 931.5 MeV/  $c^2$

Hence Q = 4.934 MeV

Kinetic energy of the  $\alpha -$ particle =  $\frac{Massnumberafterdecay}{Massnumberbeforedecay}$  $\times Q$ =  $\frac{222}{226}$  $\times$ 4.934= 4.85 MeV

$\alpha -$ particle decay of  ${}_{86}{}^{220}Rn$ emits a helium nucleus. As a result, its mass number reduces to (220-4) 216 and its atomic number reduces to (86-2) 84.

${}_{86}{}^{220}Rn$  $\to$  ${}_{84}{}^{216}Po$ +  ${}_2{}^{4}He$

Q value of emitted  $\alpha -$ particle = (Sum of initial mass – Sum of final mass)  $\times c^2$ , where

c = Speed of light.

It is given that

m (  ${}_{86}{}^{222}Rn)$ = 220.01137 u

m (  ${}_{84}{}^{216}Po)$ = 216.00189 u

m (  ${}_2{}^{4}He)$ = 4.002603 u

Q value = [(220.01137) – (216.00189 + 4.002603)]  $c^2$

= 6.877  $\times {10}^{-3}{uc}^2$

But 1 u = 931.5 MeV/  $c^2$

Hence Q = 6.406 MeV

Kinetic energy of the  $\alpha -$ particle =  $\frac{Massnumberafterdecay}{Massnumberbeforedecay}$  $\times Q$ =  $\frac{216}{222}$  $\times$ 6.406= 6.23 MeV

 

Q.13.13 The radionuclide  ${}_6{}^{11}\mathit{C}$ decays according to

${}_6{}^{11}\mathit{C}\mathit{}\to \mathit{}{}_5{}^{11}\mathit{B}+\mathit{}{\mathit{e}}^{+}+\mathit{}\mathit{v}\mathit{}:\mathit{}{\mathit{T}}_{1/2}$ =20.3 min

The maximum energy of the emitted positron is 0.960 MeV.

Given the mass values:

m (  ${}_6{}^{11}\mathit{C})$ = 11.011434 u and m (  ${}_5{}^{11}\mathit{B}$ ) = 11.009305 u,

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

Ans.13.13 The given values are

m (  ${}_6{}^{11}C)$ = 11.011434 u and m (  ${}_5{}^{11}B$ ) = 11.009305 u

The given nuclear reaction:

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

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

The maximum energy possessed by the emitted positron = 0.960 MeV

The change in the Q-value (ΔQ) of the nuclear masses of the  ${}_6{}^{11}C$

ΔQ =  $\left[m\text{'}({}_6{}^{11}C)-\{m\text{'}\left({}_5{}^{11}B\right)+{\mathrm{m}}_{\mathrm{e}}\}\mathrm{}\right]c^2\dots \dots \dots \dots \dots \left(1\right)$

where

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

c = speed of the light

m’ = Respective nuclear masses

If atomic masses are used instead of nuclear masses, then we have to add 6  $m_e$ in the case of  ${}_6{}^{11}C$ and 5  $m_e$ in the case of  ${}_5{}^{11}B$ .

Hence the equation (1) reduces to

ΔQ =  $\left[m({}_6{}^{11}C)-m\left({}_5{}^{11}B\right)-{2\mathrm{m}}_{\mathrm{e}}\right]c^2$

=  $\left[11.011434-11.009305-2\times 0.000548\right]c^2$ u

=1.033  $\times {10}^{-3}{c}^2$ u

But 1 u = 931.5 MeV/  $c^2$

ΔQ = 0.962 MeV

The value of  $\Delta Q$ is almost comparable to the maximum energy of the emitted positron.

 

Q.13.14 The nucleus  ${}_{10}{}^{23}\mathit{N}\mathit{e}$ decays by  ${\mathit{\beta }}^{-}$ emission. Write down the  $\mathit{\beta }$ -decay equation and determine the maximum kinetic energy of the electrons emitted. Given that:

m (  ${}_{10}{}^{23}\mathit{N}\mathit{e})$ = 22.994466 u

m (  ${}_{11}{}^{23}\mathit{N}\mathit{a})$ = 22.989770 u.

Ans.13.14 In  ${\beta }^{-}$ emission, the number of protons increase by 1 and one electron and an antineutrino are emitted from the parent nucleus.

${\beta }^{-}$ emission of the nucleus  ${}_{10}{}^{23}Ne$ :

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

It is given that:

Atomic mass m (  ${}_{10}{}^{23}Ne)$ = 22.994466 u

Atomic mass m (  ${}_{11}{}^{23}Na)$ = 22.989770 u

Mass of an electron,  $m_e$ = 0.000548 u

Q value of the given reaction is given as Q =  $\left[m\left({}_{10}{}^{23}Ne\right)-\left\{m\right({}_{11}{}^{23}Na)+m_e\}\right]c^2$

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

Therefore Q = {22.994466 - 22.989770}  $c^2$ = 4.696  $\times {10}^{-3}{c}^2$ u

But 1 u = 931.5 MeV/  $c^2$

Q = 4.374 MeV

The daughter nucleus is too heavy as compared to  $e^{-}$ and  $\bar{\nu }$ . Hence, it carries negligible energy. The kinetic energy of the antineutrino is nearly zero. Hence, the maximum kinetic energy of the emitted electrons is almost equal to the Q-value, i.e. 4.374 MeV.

 

Q.13.15 The Q value of a nuclear reaction A + b  $\to \mathit{}$ C + d is defined by

Q = [  ${\mathit{m}}_{\mathit{A}}+\mathit{}{\mathit{m}}_{\mathit{b}}$ –  ${\mathit{m}}_{\mathit{C}}$ –  ${\mathit{m}}_{\mathit{d}}$ ]  ${\mathit{c}}^2$

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

(ii)  ${}_6{}^{12}\mathit{C}$ +  ${}_6{}^{12}\mathit{C}$  $\to$  ${}_{10}{}^{20}\mathit{N}\mathit{e}$ +  ${}_2{}^4\mathit{H}\mathit{e}$

Atomic masses are given to be

m (  ${}_1{}^2\mathit{H}$ ) = 2.014102 u

m (  ${}_1{}^3\mathit{H}$ ) = 3.016049 u

m (  ${}_6{}^{12}\mathit{C}$ ) = 12.000000 u

m (  ${}_{10}{}^{20}\mathit{N}\mathit{e}$ ) = 19.992439 u

Ans.13.15 The given nuclear reaction is

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

Atomic mass

m (  ${}_1{}^{1}H$ ) = 1.007825 u

m (  ${}_1{}^{2}H$ ) = 2.014102 u

m (  ${}_1{}^{3}H$ ) = 3.016049 u

The Q-value of the reaction can be written as:

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

=  $\left[1.007825+\mathrm{}3.016049-2\times 2.014102\right]c^2$

= (-4.33  $\times {10}^{-3}$ )  $c^2$

But 1 u = 931.5 MeV/  $c^2$

Q = -4.0334 MeV

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

The given nuclear reaction is

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

Atomic mass

m (  ${}_6{}^{12}C$ ) = 12.000000 u

m (  ${}_{10}{}^{20}Ne$ ) = 19.992439 u

m (  ${}_2{}^{4}He$ ) = 4.002603 u

The Q-value of this reaction is given as:

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

=  $\left[2\times 12.0-19.992439-4.002603\right]c^2$

=4.958  $\times {10}^{-3}{c}^2$ u

=4.958  $\times {10}^{-3}\times 931.5$

=4.6183 MeV

The positive sign shows that the reaction is exothermic.

 

Q.13.16 Suppose, we think of fission of a  ${}_{26}{}^{56}\mathit{F}\mathit{e}$ nucleus into two equal fragments,  ${}_{13}{}^{28}\mathit{A}\mathit{l}$ . Is the fission energetically possible? Argue by working out Q of the process. Given

m (  ${}_{26}{}^{56}\mathit{F}\mathit{e})$ = 55.93494 u and m (  ${}_{13}{}^{28}\mathit{A}\mathit{l})$ = 27.98191 u.

Ans.13.16 The fission can be shown as:

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

It is given that atomic mass

m ( ${}_{26}{}^{56}Fe)$= 55.93494 u

m ( ${}_{13}{}^{28}Al)$= 27.98191 u

The Q-value of this reaction is given as:

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

= $\left[55.93494\mathrm{}-2\times 27.98191\right]c^2$

= -0.02888 $c^2$

= -0.02888 $\times 931.5$  MeV

= - 26.902 MeV

The Q value of the fission is negative, therefore the fission is not possible energetically. Q value needs to be positive for a fission.

 

Q. 13.17 The fission properties of ${}_{94}{}^{239}\mathit{P}\mathit{u}$  are very similar to those of ${}_{92}{}^{235}\mathit{U}$ . The average energy released per fission is 180 MeV. How much energy, in MeV, is released if all the atoms in 1 kg of pure ${}_{94}{}^{239}\mathit{P}\mathit{u}$undergo fission?

Ans.13.17 The average energy released per fission of ${}_{94}{}^{239}Pu$   $E_{avg}$= 180 MeV

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

Avogadro’s number, $\mathrm{<br>}$ $N_A$ = 6.023 $\times {10}^{23}$

Mass number of ${}_{94}{}^{239}\mathit{P}\mathit{u}$  = 239 gm

Hence, number of atoms in 1000 g  ${}_{94}{}^{239}Pu$ , N = $\frac{6.023\times {10}^{23}}{239}$ $\times 1000$= 2.52 $\times {10}^{24}$

Total energy released during the fission of 1 kg of   ${}_{94}{}^{239}Pu$ , E = $E_{avg}$ $\times$N

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

 

Q.13.18 A 1000 MW fission reactor consumes half of its fuel in 5.00 y. How much ${}_{92}{}^{235}\mathit{U}$ $<br>$ did it contain initially? Assume that the reactor operates 80% of the time, that all the energy generated arises from the fission of  ${}_{92}{}^{235}\mathit{U}$ and that this nuclide is consumed only by the fission process.

Ans.13.18 Half life of the fuel in the fission reactor,  = 5 years = 5 $\times 365\times 24\times 60\times 60s$

= 157.68 $\times {10}^{6}s$

We know that in the fission of 1 g of ${}_{92}{}^{235}\mathit{U}$ $<br>$ , the energy released = 200 MeV

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

Therefore 1 gm of ${}_{92}{}^{235}\mathit{U}$ $<br>$ contains = $\frac{6.023\times {10}^{23}}{235}$= 2.563 $\times {10}^{21}$ atoms

The total energy Q generated per gm of   ${}_{92}{}^{235}U$ = 200 $\times$2.563 $\times {10}^{21}$ MeV/g = 5.126 $\times {10}^{23}$  $<br>$ MeV/g = 5.126 $\times {10}^{23}\times 1.6\times {10}^{-19}\times {10}^6$ J/g = 8.20 $\times {10}^{10}$  $<br>$ J/g

Since the reactor operates only 80% of the time, hence the amount of ${}_{92}{}^{235}\mathit{U}$ $<br>$ in 5 years is given by $\frac{0.8\times 157.68\times {10}^6\times 1000\times {10}^6}{8.20\times {10}^{10}}$ = 1538 kg

Hence, initial amount of fuel = 2 $\times 1538kg$ = 3076 kg

 

Q.13.19 How long can an electric lamp of 100W be kept glowing by fusion of 2.0 kg of deuterium? Take the fusion reaction as

${}_1{}^2\mathit{H}\mathit{}$  +  ${}_1{}^2\mathit{H}\mathit{}$  $\to$ ${}_2{}^3\mathit{H}\mathit{e}$ + n+3.27 MeV

Ans.13.19 The given fusion reaction is

$\mathit{}$  ${}_1{}^{2}H$+ ${}_1{}^{2}H$  $\to$ ${}_2{}^3\mathit{H}\mathit{e}$ + n+3.27 MeV

Amount of deuterium, m = 2 kg

1 mole, i.e. 2 g of deuterium contains 6.023  atoms

Hence 2 kg of deuterium contains = 
$\frac{6.023\times {10}^{23}}{2}$ $\times 2\times {10}^3$atoms = 6.023 $\times {10}^{26}$ atoms

It can be inferred from the fission reaction that when 2 atoms of deuterium fuse, 3.27 MeV of energy is released.

Hence total energy released from 2 kg of deuterium, E = 
$\frac{3.27}{2}$ $\times$6.023 $\times {10}^{26}$  MeV

=   $\frac{3.27}{2}$ $\times$ 6.023 $\times {10}^{26}\times {10}^6\times 1.6\times {10}^{-19}$ J = 1.5756 $\times {10}^{14}$  $<br>$ J

Power of the electric bulb, P = 100 W = 100 J/s

Hence energy consumed by the bulb per second = 100 J

Therefore, total time the electric bulb will glow = $\frac{1.5756\times {10}^{14}}{100}$ seconds = 1.5756 $\times {10}^{12}$secs

= 49.96 $\times {10}^3$ years

 

Q.13.20 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 fm.)

Ans.13.20 When two deuterons collide head-on, the distance between their centers, d is given as

d = Radius of 1st deuteron + Radius of 2nd deuteron

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

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

Charge on a deuteron nucleus = Charge on an electron = 1.6 $\times {10}^{-19}$ C

Potential energy of two-deuteron system:  $<math>\; <mi>\; V\; </mi>\; </math>$ = $\frac{e^2}{4\pi {ϵ}_{0}d}$ $$

Where ${ϵ}_0$= permittivity of free space

 = 9 $\times {10}^9$ N $m^2{C}^{-1}$

V = $\frac{{(1.6\times {10}^{-19})}^2\times 9\mathrm{}\times {10}^9}{4\mathrm{}\times {10}^{-15}}$ J = $\frac{{(1.6\times {10}^{-19})}^2\times 9\mathrm{}\times {10}^9}{4\mathrm{}\times {10}^{-15}\times 1.6\times {10}^{-19}}$ eV = 360 $\times {10}^3$ $\frac{{<br>}^{}}{}$ eV = 360 keV

Hence the height of the potential barrier of the two-deuteron system is 360 keV.

 

Q.13.21 From the relation R =  , where ${\mathit{R}}_0$  is a constant and A is the mass number of a nucleus, shows that the nuclear matter density is nearly constant (i.e. independent of A).

Ans.13.21 We have the expression for nuclear radius as:

R = $R_0{A}^{1/3}$

Where $R_0$ = constant

A = mass number of nucleus

Let m be the average mass of the nucleus, hence mass of the nucleus = mA

Nuclear matter density $\rho$ can be written as

  $\rho =\frac{Massofthenucleus}{Volumeofthenucleus}$ =  $\frac{mA}{\frac{4}{3}\pi R^3}$ =  $\frac{3mA}{4\pi {\left(R_0{A}^{\frac{1}{3}}\right)}^3}$ =  $\frac{3mA}{4\pi R_0^{3}A}$ = $\frac{3m}{4\pi R_0^3}$

Hence, the nuclear mass density is independent of A. It is nearly constant

 

Q.13.22 For the ${\mathit{\beta }}^{+}$ (positron) emission from a nucleus, there is another competing process known as electron capture (electron from an inner orbit, say, the K–shell, is captured by the nucleus and a neutrino is emitted).

${\mathit{<br>}}^{}$  ${\mathit{e}}^{+}$ +  ${}_{\mathit{Z}}{}^{\mathit{A}}\mathit{X}$ $\to \mathit{}{}_{\mathit{z}-1}{}^{\mathit{A}}\mathit{Y}+\mathit{v}$

Show that if ${\mathit{\beta }}^{++}$emission is energetically allowed, electron capture is necessarily allowed but not vice–versa.

Ans.13.22 Let the amount on energy released during the electron capture process be $Q_1$ . The nuclear reaction can be written as:

  $e^{+}$+ ${}_Z{}^{A}X$ $\to {}_{z-1}{}^{A}Y+v+Q_1$ ……………………..(1)

Let the amount of energy released during the positron capture process be $Q_2$. The nuclear reaction can be written as:

  ${}_Z{}^{A}X$ $\to {}_{z-1}{}^{A}Y+e^{+}+v$ + $Q_2$ …………………….(2)

Let us assume

$m_N\left({}_Z{}^{A}X\right)$ = Nuclear mass of ${}_Z{}^{A}X$

$m_N\left({}_Z{}^{A}X\right)$ = Atomic mass of ${}_Z{}^{A}X$

$m_N\left({}_{Z-1}{}^{A}Y\right)$ = Nuclear mass of ${}_{Z-1}{}^{A}Y$

$m_N\left({}_{Z-1}{}^{A}Y\right)$ = Atomic mass of ${}_{Z-1}{}^{A}Y$

$m_e$ = mass of the electron

c = speed of light

 

Q-value of the electron capture reaction is given as:

${\mathrm{<br>}}_{}$ $Q_1$ = [ 

= [ $m\left({}_Z{}^{A}X\right)-Z{m}_e+m_e-m\left({}_{Z-1}{}^{A}Y\right)+(Z-1)m_e]c^2$ ${\mathrm{<br>}}_{}$

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

Q-value of the positron capture reaction is given as:

$\mathrm{<br>}$ $Q_2$ = [ $m_N\left({}_Z{}^{A}X\right)-m_N\left({}_{Z-1}{}^{A}Y\right)-m_e]c^2$ ${\mathrm{<br>}}_{}$

= [ $m\left({}_Z{}^{A}X\right)-Z{m}_e-m\left({}_{Z-1}{}^{A}Y\right)+\left(Z-1\right)m_e-m_e]c^2$ ${\mathrm{<br>}}_{}$

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

It can be inferred that if $Q_2>0$ , then $Q_1>0$ ${\mathrm{<br>}}_{}$ , also if  means that $Q_2>0$.

In other words, this means that if ${\beta }^{+}$ emission is energetically allowed, then the electron capture process is necessarily allowed, but not vice-versa. This is because Q-value must be positive for an energetically-allowed nuclear reaction.

 

Q.13.23 In a periodic table the average atomic mass of magnesium is given as 24.312 u. The average value is based on their relative natural abundance on earth. The three isotopes and their masses are ${}_{12}{}^{24}\mathit{M}\mathit{g}$ (23.98504u), ${}_{12}{}^{25}\mathit{M}\mathit{g}$ $\mathit{<br>}_{\mathrm{<br>}}$(24.98584u) and ${}_{12}{}^{26}\mathit{M}\mathit{g}$ (25.98259u). The natural abundance of  ${}_{12}{}^{24}\mathit{M}\mathit{g}$ is 78.99% by mass. Calculate the abundances of other two isotopes.

Ans.13.23 Average atomic mass of magnesium, m = 24.312 u

Mass of magnesium isotope  ${}_{12}{}^{24}\mathit{M}\mathit{g}$ ,  = 23.98504u

Mass of magnesium isotope   ${}_{12}{}^{25}Mg$ , $m_2$= 24.98584u

Mass of magnesium isotope   ${}_{12}{}^{26}Mg$ , $m_3$= 25.98259u

Let the abundance of magnesium isotope   ${}_{12}{}^{24}Mg$ be ${\eta }_1$ = 78.99 %

Let the abundance of magnesium isotope   ${}_{12}{}^{25}Mg$ be ${\eta }_2$= x %

Therefore, the abundance of magnesium isotope   ${}_{12}{}^{26}Mg$ be ${\eta }_3$= (100 - 78.99 - x) %

= (21.01 – x)%

The average atomic mass can be expressed as:

m = $\frac{m_1{\eta }_1+m_2{\eta }_2+m_3{\eta }_3}{{\eta }_1+{\eta }_2+{\eta }_3}$ =  $\frac{23.98504\times 78.99+24.98584x+25.98259(21.01-x)}{100}$ =  $\frac{1894.578+24.98584x+545.894-25.98259x}{100}$= $\frac{2440.472-0.99675x}{100}$

24.312 = $\frac{2440.472-0.99675x}{100}$

x = 9.3%

Therefore the abundance of ${}_{12}{}^{25}Mg$ $\frac{\mathrm{<br>}}{}$ is 9.3% and the abundance of  ${}_{12}{}^{26}Mg$ is (21.01 – 9.3)11.71%

 

Q.13.24 The neutron separation energy is defined as the energy required to remove a neutron from the nucleus. Obtain the neutron separation energies of the nuclei ${}_{20}{}^{41}\mathit{C}\mathit{a}$  and  from the following data:

m( ${}_{20}{}^{40}\mathit{C}\mathit{a})$  = 39.962591 u

m( ${}_{20}{}^{41}\mathit{C}\mathit{a})$ = 40.962278 u

m(  ${}_{13}{}^{26}\mathit{A}\mathit{l})$ = 25.986895 u

m( ${}_{13}{}^{27}\mathit{A}\mathit{l})$ = 26.981541 u

Ans.13.24 If a neutron ${}_0{}^{1}n)$  is removed from ${}_{20}{}^{41}Ca$ , the corresponding reaction can be written as:

${}_{20}{}^{41}Ca\to$ ${}_{20}{}^{40}Ca$ + ${}_0{}^{1}n$

The separation energies are

For ${}_{20}{}^{41}Ca$  : Separation energy = 8.363007 MeV

For ${}_{13}{}^{27}Al$  : Separation energy = 13.059 MeV

It is given that

m( ${}_{20}{}^{40}Ca)$  = 39.962591 u

m( ${}_{20}{}^{41}Ca)$ = 40.962278 u

m( ${}_0{}^{1}n)$ = 1.008665 u

 

The mass defect of the reaction is given as:

Δm = m ${}_{20}{}^{40}Ca)+$ (  m( ${}_0{}^{1}n)-$ m( ${}_{20}{}^{41}Ca)$

= 39.962591 + 1.008665 - 40.962278

= 8.978 $\times {10}^3$ u

=8.363007 MeV

 

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

${}_{13}{}^{27}Al\to$ ${}_{13}{}^{26}Al$ + ${}_0{}^{1}n$

It is given that

m( ${}_{13}{}^{26}Al)$ = 25.986895 u

m( ${}_{13}{}^{27}Al)$  = 26.981541 u

m( ${}_0{}^{1}n)$ = 1.008665 u

The mass defect of the reaction is given as:

Δm = m( ${}_{13}{}^{26}Al)+$  m( ${}_0{}^{1}n)-$ m( ${}_{13}{}^{27}Al)$

= 25.986895 + 1.008665 - 26.981541

= 0.014019 u

=13.0586985 MeV

 

Q.13.25 A source contains two phosphorous radio nuclides ${}_{15}{}^{32}P{,T}_{1/2}$ = 14.3d and ${}_{15}{}^{33}P,T_{1/2}$ = 25.3d. Initially, 10% of the decays come from ${}_{15}{}^{33}P$ . How long one must wait until 90% do so?

Ans.13.25 Half life of ${}_{15}{}^{32}P{,T}_{1/2}$ = 14.3 days

Half life of ${}_{15}{}^{33}P,T_{1/2}$  = 25.3days and ${}_{15}{}^{33}P$  decay is 10% of the total amount of decay

The source has initially 10 % of ${}_{15}{}^{33}P$ 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:

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, the number ratio, $\frac{N\text{'}}{9N}$  = ${\left(\frac{1}{2}\right)}^{t/T_{1/2}}$

N’ = 9N ${\left(2\right)}^{-t/14.3}$ …………….(1)

 

For ${}_{15}{}^{33}P$  nucleus, the number ratio, $\frac{9N\text{'}}{N}$ = ${\left(\frac{1}{2}\right)}^{t/T_{1/2}}$

9N’ = N ${\left(2\right)}^{-t/25.3}$  …………….(2)

 

On dividing equation (1) by equation (2), we get:

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

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

Taking log on both sides

log 1 – log 81 = $\frac{-11t}{25.3\times 14.3}$ log2

0 – 1.908 = ( $\frac{-11t}{361.79}$ ) $\times 0.301$

t = 208.5 days

Hence, it will take about 208.5 days for 90% decay of $P_{15}^{33}$.

 

Q.13.26 Under certain circumstances, a nucleus can decay by emitting a particle more massive than an $\mathit{\alpha }-$ particle. Consider the following decay processes:

${}_{88}{}^{223}\mathit{R}\mathit{a}\mathit{}\to {}_{82}{}^{209}\mathit{P}\mathit{b}+\mathit{}{}_6{}^{14}\mathit{C}$

 

${}_{88}{}^{223}\mathit{R}\mathit{a}\mathit{}\to {}_{86}{}^{219}\mathit{R}\mathit{n}+\mathit{}{}_2{}^4\mathit{H}\mathit{e}$

Calculate the Q-values for these decays and determine that both are energetically allowed.

Ans.13.26 For the emission of ${}_6{}^{14}C$ , the nuclear reaction is:

${}_{88}{}^{223}Ra\to {}_{82}{}^{209}Pb+{}_6{}^{14}C$

We know that:

Mass of ${}_{88}{}^{223}Ra$ , $m_1$ = 223.01850 u

Mass of ${}_{82}{}^{209}Pb$ , $m_2$  = 208.98107 u

Mass of ${}_6{}^{14}C$ , $m_3$ = 14.00324 u

 

Hence, the Q-value of the reaction is given as:

Q = ( $m_1$ - $m_2$  - $m_3$ ) $c^2$

= (223.01850 - 208.98107 - 14.00324) $c^2$ u

= 0.03419 $c^2$ u

= 0.03419 MeV = 31.848 MeV

Hence, the Q-value of the nuclear reaction is 31.848 MeV, since the value is positive, the reaction is energetically allowed.

 

For the emission of ${}_2{}^{4}He$ , the nuclear reaction is:

${}_{88}{}^{223}Ra\to {}_{86}{}^{219}Rn+{}_2{}^{4}He$

We know that:

Mass of ${}_{88}{}^{223}Ra$ , $m_1$ = 223.01850 u

Mass of ${}_{86}{}^{219}Rn$ , $m_2$  = 219.00948 u

Mass of ${}_2{}^{4}He$ , $m_3$ = 4.00260 u

 

Hence, the Q-value of the reaction is given as:

Q = ( $m_1$ - $m_2$  - $m_3$ ) $c^2$

= (223.01850 - 219.00948 - 4.00260) $c^2$ u

= 6.42 $\times {10}^{-3}{c}^2$ u

= 6.42 $\times {10}^{-3}\times 931.5$ MeV = 5.980 MeV

Hence, the Q-value of the nuclear reaction is 5.98 MeV, since the value is positive, the reaction is energetically allowed.

 

Q.13.27 Consider the fission of ${}_{92}{}^{238}U$ by fast neutrons. In one fission event, no neutrons are emitted and the final end products, after the beta decay of the primary fragments, are ${}_{58}{}^{140}\mathit{C}\mathit{e}$ and ${}_{44}{}^{99}\mathit{R}\mathit{u})$   . Calculate Q for this fission process. The relevant atomic and particle masses are

m( ${}_{92}{}^{238}\mathit{U})$ =238.05079 u

m( ${}_{58}{}^{140}\mathit{C}\mathit{e}$ ) =139.90543 u

m( ${}_{44}{}^{99}\mathit{R}\mathit{u})$  = 98.90594 u

 Ans.13.27 In the fission of ${}_{92}{}^{238}U$ , ${\beta }^{-}$ 10  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 given that:

Mass of a ${}_{92}{}^{238}\mathrm{U}\mathrm{}\mathrm{n}\mathrm{u}\mathrm{c}\mathrm{l}\mathrm{e}\mathrm{u}\mathrm{s},\mathrm{}{\mathrm{m}}_{1\mathrm{}}$ = 238.05079 u

Mass of a ${}_{58}{}^{140}\mathrm{C}\mathrm{e}\mathrm{}\mathrm{n}\mathrm{u}\mathrm{c}\mathrm{l}\mathrm{e}\mathrm{u}\mathrm{s},\mathrm{}{\mathrm{m}}_2$ =139.90543 u

Mass of a ${}_{44}{}^{99}\mathrm{R}\mathrm{u}\mathrm{}\mathrm{n}\mathrm{u}\mathrm{c}\mathrm{l}\mathrm{e}\mathrm{u}\mathrm{s},\mathrm{}{\mathrm{m}}_3$ = 98.90594 u

Mass of a neutron ${}_0{}^{1}n$ , $m_4$ = 1.00865 u

 

Q value of the above equation,

Q = $\left[m^{\text{'}}\left({}_{92}{}^{238}\mathrm{U}\right)+\mathrm{}\mathrm{m}\mathrm{\text{'}}({}_0{}^1\mathrm{n})-{\mathrm{m}}^{\mathrm{\text{'}}}({}_{58}{}^{140}Ce)-m^{\text{'}}({}_{44}{}^{99}Ru)-10m_e\right]c^2$

Where

m’ = represents the corresponding atomic masses of the nuclei.

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

m’( ${}_{58}{}^{140}Ce)$ = ${\mathrm{m}}_2-\mathrm{}58m_e$

m’( ${}_{44}{}^{99}Ru)$  = $m_3-44m_e$

m’( ${}_0{}^{1}n)=m_4$

Substituting these values, we get

Q = $\left[{\mathrm{m}}_{1\mathrm{}}-\mathrm{}92m_e+m_4-{\mathrm{m}}_2+\mathrm{}58m_e-m_3+44m_e-10m_e\right]c^2$

= $\left[{\mathrm{m}}_{1\mathrm{}}+\mathrm{}{m}_4-m_3-{\mathrm{m}}_2\right]c^2$

= $\left[238.05079+1.00865-98.90594-139.90543\right]c^2$ u

=0.24807 $c^2$ u

= 231.077 MeV

 

 

Q.13.28 Consider the D–T reaction (deuterium–tritium fusion)

${}_1{}^2\mathit{H}+\mathit{}{}_1{}^3\mathit{H}\mathit{}\to \mathit{}{}_2{}^4\mathit{H}\mathit{e}+\mathit{n}$

 (a) Calculate the energy released in MeV in this reaction from the data:

m( ${}_1{}^2\mathit{H})$  =2.014102 u

m( ${}_1{}^3\mathit{H})$  =3.016049 u

(b) Consider the radius of both deuterium and tritium to be approximately 2.0 fm. 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(3kT/2); k = Boltzmann’s constant, T = absolute temperature.)

Ans.13.28 The equation for deuterium-tritium fusion is given as:

${}_1{}^{2}H+{}_1{}^{3}H\to {}_2{}^{4}He+n$

 It is given that

Mass of ( ${}_1{}^2\mathrm{H}),\mathrm{}{\mathrm{m}}_1$  = 2.014102 u

Mass of ( ${}_1{}^3\mathrm{H})$ , ${\mathrm{m}}_2=\mathrm{}$ 3.016049 u

Mass of ( ${}_2{}^{4}He),m_3$  = 4.002603 u

Mass of ( ${}_0{}^{1}n),m_4$ = 1.008665 u

Q-value of the given D-T reaction is:

Q = $\left[{\mathrm{m}}_1+{\mathrm{m}}_2-m_3-m_4\right]c^2$

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

= 0.018883 $c^{2}u$

= 17.59 MeV

 

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

Distance between the centers of the nucleus when they touch each other,

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

Charge on the deuterium and tritium nucleus = e

Hence the repulsive potential energy between the two nuclei is given as:

V = $\frac{e^2}{4\pi {ϵ}_{0}d}$

Where,

${ϵ}_0$  = permittivity of free space

 It is given that $\frac{1}{4\pi {ϵ}_0}$  = 9 $\times {10}^9$ N $m^2{C}^{-2}$

Hence, V = $\frac{{(1.6\times {10}^{-19})}^2}{4\times {10}^{-15}}$ $\times$  9 $\times {10}^9$ = 5.76 $\times {10}^{-14}$  J = $\frac{5.76\times {10}^{-14}}{1.6\times {10}^{-19}}$  eV = 360 keV

Hence, 360 keV of kinetic energy is needed to overcome the Coulomb repulsion between the two nuclei.

It is given that

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

Where

k = Boltzmann constant = 1.38 $\times {10}^{-23}$ $m^{2}kg{s}^{-2}{K}^{-1}$

T = Temperature required to trigger 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

Hence the gas must be heated to 1.39 $\times {10}^9$ K to initiate the reaction.

 

Q.13.29 Obtain the maximum kinetic energy of $\mathit{\beta }$ -particles, and the radiation frequencies of  decays in the decay scheme shown in Fig. 13.6. You are given that

m( ${}_{}{}^{198}\mathit{A}\mathit{u}$ ) = 197.968233 u

m( ${}_{}{}^{198}\mathit{H}\mathit{g}$ ) =197.966760 u

Ans.13.29 It can be observed from the given $\gamma -$ decay diagram that ${\gamma }_1$ decays from 1.088 MeV energy level to the 0 MeV energy level.

Hence the energy corresponding to  decay is given as:

 = 1.088 – 0 = 1.088 MeV = 1.088 $\times {10}^6$  eV = 1.088 $\times {10}^6\times 1.6\times {10}^{-19}$ J

= 1.7408 $\times {10}^{-13}$ J

We know, $E_1=h{\nu }_1$ , where

${\nu }_1$ = Frequency of radiation radiated by ${\gamma }_1$  decay

$h=\mathrm{P}\mathrm{l}\mathrm{a}\mathrm{n}\mathrm{c}{\mathrm{k}}^{\mathrm{\text{'}}}\mathrm{s}\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{t}$ = 6.6 $\times {10}^{-34}$ Js

 Hence, ${\nu }_1$ = $\frac{E_1}{h}$  = $\frac{1.7408\times {10}^{-13}}{6.6\times {10}^{-34}}$  = 2.637 $\times {10}^{20}$  Hz

 

• It can be observed from the given $\gamma -$ decay diagram that ${\gamma }_2$ decays from 0.412 MeV energy level to the 0 MeV energy level.

Hence the energy corresponding to ${\gamma }_2$  decay is given as:

$E_2$ = 0.412 – 0 = 0.412 MeV = 0.412 $\times {10}^6$  eV = 0.412 $\times {10}^6\times 1.6\times {10}^{-19}$  J

= 6.592 $\times {10}^{-14}$ J

We know, $E_2=h{\nu }_2$ , where

${\nu }_2$ = Frequency of radiation radiated by ${\gamma }_2$ decay

$h=\mathrm{P}\mathrm{l}\mathrm{a}\mathrm{n}\mathrm{c}{\mathrm{k}}^{\mathrm{\text{'}}}\mathrm{s}\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{t}$ = 6.6 $\times {10}^{-34}$ Js

 Hence, ${\nu }_2$  = $\frac{E_2}{h}$ = $\frac{6.592\times {10}^{-14}}{6.6\times {10}^{-34}}$  = 9.987 $\times {10}^{19}$  Hz

• It can be observed from the given $\gamma -$ decay diagram that ${\gamma }_3$ decays from 1.008 MeV energy level to the 0.412 MeV energy level.

Hence the energy corresponding to  decay is given as:

 = 1.088 - 0.412 = 0.676 MeV = 0.676 $\times {10}^6$  eV = 0.676 $\times {10}^6\times 1.6\times {10}^{-19}$  J

= 1.0816 $\times {10}^{-13}$  J

We know, $E_3=h{\nu }_3$ , where

${\nu }_3$ = Frequency of radiation radiated by ${\gamma }_3$  decay

$h=\mathrm{P}\mathrm{l}\mathrm{a}\mathrm{n}\mathrm{c}{\mathrm{k}}^{\mathrm{\text{'}}}\mathrm{s}\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{t}$ = 6.6 $\times {10}^{-34}$  Js

 Hence, ${\nu }_3$  = $\frac{E_3}{h}$  = $\frac{1.0816\times {10}^{-13}}{6.6\times {10}^{-34}}$ = 1.639 $\times {10}^{20}$ Hz

 

Mass of ( ${}_{79}{}^{198}\mathrm{A}\mathrm{u}$ ) = 197.968233 u

Mass of ( ${}_{79}{}^{198}\mathrm{A}\mathrm{u}$ ) =197.966760 u

Energy of the highest level is given as:

E = $\left[\mathrm{m}\left({}_{79}{}^{198}\mathrm{A}\mathrm{u}\right)-\mathrm{m}\left({}_{80}{}^{198}Hg\right)\mathrm{}\right]c^2$

= $\left[197.968233-197.966760\right]c^2$ u

= 1.473 $\times {10}^{-3}{c}^2$ u

= 1.473 $\times {10}^{-3}\times 931.5$  MeV

= 1.3720995 MeV

${\beta }_1^{-}\mathrm{d}\mathrm{e}\mathrm{c}\mathrm{a}\mathrm{y}\mathrm{s}$  from 1.3720995 MeV level to 1.088 MeV level

Hence maximum kinetic energy of the ${\beta }_1^{-}$ particles = 1.3720995 - 1.088 = 0.2840995 MeV

${\beta }_2^{-}$ from 1.3720995 MeV level to 0.412 MeV level

Hence maximum kinetic energy of the ${\beta }_2^{-}$  particles = 1.3720995 – 0.412 = 0.9600995 MeV

 

Q.13.30 Calculate and compare the energy released by a) fusion of 1.0 kg of hydrogen deep within Sun and b) the fission of 1.0 kg of ${}_{92}{}^{235}U$  in a fission reactor.

 Ans.13.30 Amount of hydrogen, m = 1 kg = 1000 g

1 mole of hydrogen, i.e. 1 g of hydrogen ( ${}_1{}^{1}H)$ contains 6.023 $\times {10}^{23}$  atoms

1 kg of hydrogen contains = 1000 $\times$ 6.023 $\times {10}^{23}$  atoms = 6.023 $\times {10}^{26}$ atoms

Within Sun, four ( ${}_1{}^{1}H)$  nuclei combine and forms 1 ${}_2{}^{4}He$  nucleus. In this process 26 MeV of energy is released.

Hence the energy released from fusion of 1 kg of ${}_1{}^{1}H$  is:

$E_1$  = $\frac{6.023\times {10}^{26}\times 26}{4}$  MeV = 3.91495 $\times {10}^{27}$  MeV

 

• Amount of ${}_{92}{}^{235}U$ , m = 1 kg = 1000 g

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

1 kg of ${}_{92}{}^{235}U$ contains = $\frac{1000}{235}$ $\times$  6.023 $\times {10}^{23}$ atoms = 2.563 $\times {10}^{24}$  atoms

It is known that the amount of energy released in the fission of 1 atom of ${}_{92}{}^{235}U$  is 200 MeV.

Hence the energy released from fusion of 1 kg of ${}_{92}{}^{235}U$  is:

 = 2.563 $\times {10}^{24}\times 200$  MeV = 5.126 $\times {10}^{26}$  MeV

 

$\frac{E_1}{E_2}$ = $\frac{3.91495\times {10}^{27}}{5.126\times {10}^{26}}$ = 7.64

 

Therefore, the energy released in the fusion of 1 kg of hydrogen is nearly 8 times of the energy release during the fission of 1 kg of ${}_{92}{}^{235}U$ .

 

Q.13.31 Suppose India had a target of producing by 2020 AD, 200,000 MW of electric power, ten percent of which was to be obtained from nuclear power plants. Suppose we are given that, on an average, the efficiency of utilization (i.e. conversion to electric energy) of thermal energy produced in a reactor was 25%. How much amount of fissionable uranium would our country need per year by 2020? Take the heat energy per fission of ${}_{}{}^{235}U$  to be about 200MeV.

Ans.13.31 Amount of Electric power to be generated, P = 200000 MW

10% of which to be obtained from nuclear power.

Hence, amount of nuclear power = 10% of 2 $\times {10}^5$  MW = 2 $\times {10}^4$ MW = 2 $\times {10}^4\times {10}^6$  J/s

= 2 $\times {10}^4\times {10}^6\times 60\times 60\times 24\times 365$  J/y = 6.3072 $\times {10}^{17}$ J/y

Heat energy release per fission of a ${}_{}{}^{235}U$  nucleus, E = 200 MeV

Efficiency of a reactor = 25%

So the amount of electrical energy converted from heat energy per fission = 25% of 200 MeV = 50 MeV

= 50 $\times {10}^6$ eV = 50 $\times {10}^6\times 1.6\times {10}^{-19}$ J = 8 $\times {10}^{-12}$  J

Therefore, number of atoms required per year = $\frac{6.3072\times {10}^{17}}{8\times {10}^{-12}}$  = 7.884 $\times {10}^{28}$

1 mole of ${}_{}{}^{235}U$  = 235 gm of ${}_{}{}^{235}U$  contains 6.023 $\times {10}^{23}$ atoms

Hence the mass of 7.884 $\times {10}^{28}atoms$  = $\frac{235\times {10}^{-3}}{6.023\times {10}^{23}}$ $\times$ 7.884 $\times {10}^{28}$ = 30.76 $\times {10}^3$ kg

Hence, the Uranium needed per year is 30.76 $\times {10}^3$ kg