13.1 Obtain the binding energy (in MeV) of a nitrogen nucleus $({_{7}\textrm{N}^{14}})$, given m $({_{7}\textrm{N}^{14}})$ =14.00307 u.
solution :
The $({_{7}\textrm{N}^{14}})$ nucleus contains 7 protons and 7 neutrons.
Mass defect ,
$$\Delta m = 7m_{p} + 7 m_{n} - M_{N}$$
$$\Delta m = 7 \times 1.00783 + 7 \times 1.00867 - 14.00307$$
$$\Delta m = 7.05481 + 7.06069 - 14.00307$$
$$\Delta m = 0.11243 a.m.u$$
Binding energy (B.E) of a nitrogen nucleus is
$$B.E = \Delta m \times 931.5 MeV$$
$$B.E = 0.11243 \times 931.5$$
$$B.E = 104.67 MeV$$
13.2 Obtain the binding energy of the nuclei ${_{26}\textrm{Fe}^{56}}$ and ${_{83}\textrm{Bi}^{209}}$ in units of MeV from the following data:
$m ( {_{26}\textrm{Fe}^{56}} ) = 55.934939 u$
$m ( {_{83}\textrm{Bi}^{209}} ) = 208.980388 u$
solution :
The ${_{26}\textrm{Fe}^{56}}$ nucleus contains 26 protons and (56-26) = 30 neutrons
Mass of 26 proton = 26 × 1.007825 = 26.20345 a.m.u
Mass of 30 neutrons = 30 × 1.008665 = 30.25995 a.m.u
Total mass of 56 nucleons = 26.20345 + 30.25995 = 56.46340 a.m.u
Mass of ${_{26}\textrm{Fe}^{56}}$ nucleus = 55.934939 a.m.u
Mass defect , $\Delta m$ = 56.46340 - 55.934939 = 0.528461 a.m.u
B.E of ${_{26}\textrm{Fe}^{56}}$ nucleus =$ \Delta m \times 931.5 = 0.5284611 \times 931.5 $= 492.26 MeV
B.E /Nucleon of ${_{26}\textrm{Fe}^{56}}$ = 492.26 /56 = 8.79 MeV
The ${_{83}\textrm{Bi}^{209}}$ nucleus contains 83 protons and (209-83) = 126 neutrons
Mass of 83 protons = $83 \times 1.007825 $= 83.649475 a.m.u
Mass of 126 neutrons = $126 \times 1.008665$ = 127.091790 a.m.u
Mass of 209 nucleons = 210.741265 a.m.u
Mass of ${_{83}\textrm{Bi}^{209}}$ nucleus = 208.980388 a.m.u
Mass defect ,$ \Delta m $= 210.741265 - 208.980388 = 1.760877 a.m.u
B.E of ${_{83}\textrm{Bi}^{209}}$ nucleus =$ \Delta m \times 931.5 = 1.760877 \times 931.5$ = 1640.3 MeV
B E / Nucleon of ${_{83}\textrm{Bi}^{209}}$ = 1640.3 /209 = 7.85 MeV
Clearly , ${_{26}\textrm{Fe}^{56}}$ has greater B.E per Nucleon than ${_{83}\textrm{Bi}^{209}}$
13.3 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 ${_{29}\textrm{Cu}^{63}}$ atoms (of mass 62.92960 u).
Solution :
Each copper atom has 29 protons and (63 -29) = 34 neutrons
Mass defect in each copper atom = Mass of Nucleons - Mass of Nucleus
$$\Delta m = 29 m_{p} + 34 m_{n} - M_{cu}$$
$$\Delta m = 29 \times 1.007825 + 34 \times 1.008667 - 62.92960$$
$$\Delta m = 63.52185 - 62.92960$$
$$\Delta m= 0.59225 a.m.u$$
No of atoms in 3g coin
$$N= \frac{6.023 \times 10^{23} \times 3}{63}$$
$$N= 2.868 \times 10^{22}$$
Total mass defect of all the atoms in the coin is
$$\Delta m = 0.59225 \times 2.868 \times 10^{22} = 1.6985 \times 10^{22} a.m.u$$
Energy required to separate all the neutrons and protons in the coin
$$E= \Delta m \times 931.5 = 1.6985 \times 10^{22} \times 931.5 = 1.58 \times 10^{25} MeV$$
13.4 Obtain approximately the ratio of the nuclear radii of the gold isotope ${_{79}\textrm{Au}^{197}}$ and the silver isotope ${_{47}\textrm{Ag}^{107}}$.
Solution :
Now $R = R_{0}A^{1/3}$
$$\frac{R_{Au}}{R_{Ag}} = \frac{R_{0}(197)^{1/3}}{R_{0}(107)^{1/3}}$$
$$\frac{R_{Au}}{R_{Ag}}= \left ( \frac{197}{107} \right )^{1/3}$$
$$\frac{R_{Au}}{R_{Ag}}= (1.81)^{1/3}$$
$$\frac{R_{Au}}{R_{Ag}} = 1.22$$
13.5 The Q value of a nuclear reaction $A + b \rightarrow C + d$ 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.
(i) ${_{1}\textrm{H}^{1}} + {_{1}\textrm{H}^{3}} \rightarrow {_{1}\textrm{H}^{2}} + {_{1}\textrm{H}^{2}}$
(ii) ${_{6}\textrm{C}^{12}} + {_{6}\textrm{C}^{12}} \rightarrow {_{10}\textrm{Ne}^{20}} + {_{2}\textrm{He}^{4}}$
Atomic masses are given to be :
$m{_{1}\textrm{H}^{2}} = 2.014102u$
$m{_{1}\textrm{H}^{3}} = 3.016049u$
$m({_{6}\textrm{C}^{12}}) 12.000000u$
$m{_{10}\textrm{Ne}^{20}} =19.992439u$
Solution :
(i) ${_{1}\textrm{H}^{1}} + {_{1}\textrm{H}^{3}} \rightarrow {_{1}\textrm{H}^{2}} + {_{1}\textrm{H}^{2}}$
$Q = [m({_{1}\textrm{H}^{1}}) + m({_{1}\textrm{H}^{3}}) - (m {_{1}\textrm{H}^{2}}+m {_{1}\textrm{H}^{2}})] \times 931.5 MeV$
$Q = [m({_{1}\textrm{H}^{1}}) + m({_{1}\textrm{H}^{3}}) - 2m {_{1}\textrm{H}^{2}}] \times 931.5 MeV$
$Q= [1.007825 + 3.016049 - 2 \times 2.014102 ] \times 931.5 MeV$
$Q= -0.0043330 \times 931.5 MeV$
$Q= - 4.03 MeV$
Since Q-value is negative , the reaction is endothermic
(ii) $${_{6}\textrm{C}^{12}} + {_{6}\textrm{C}^{12}} \rightarrow {_{10}\textrm{Ne}^{20}} + {_{2}\textrm{He}^{4}}$$
$$Q = [m({_{6}\textrm{C}^{12}}) +m({_{6}\textrm{C}^{12}}) - (m{_{10}\textrm{Ne}^{20}}+ m {_{2}\textrm{He}^{4}})] \times 931.5 MeV$$
$$Q= [2 \times 12.00000-19.992439 - 4.002603] \times 931.5 MeV$$
$$Q= 0.004958 \times 931.5 MeV$$
$$Q= 4.618 MeV$$
Since Q-value is positive, the reaction is exothermic.
13.6 Suppose, we think of fission of a ${_{26}\textrm{Fe}^{56}}$ nucleus into two equal fragments, ${_{13}\textrm{AI}^{28}}$ . Is the fission energetically possible? Argue by working out Q of the process. Given $m( {_{26}\textrm{Fe}^{56}} )$ = 55.93494 u and $m ( {_{13}\textrm{AI}^{28}} )$= 27.98191 u.
Solution:
$${_{26}\textrm{Fe}^{56}} \rightarrow {_{13}\textrm{AI}^{28}} +{_{13}\textrm{AI}^{28}}$$
$$Q = [m({_{26}\textrm{Fe}^{56}}) - ((m({_{13}\textrm{AI}^{28}})+(m({_{13}\textrm{AI}^{28}}))] \times 931.5 MeV$$
$$Q = [m({_{26}\textrm{Fe}^{56}}) - 2m({_{13}\textrm{AI}^{28}})] \times 931.5 MeV$$
$$Q= [55.93494 - 2 \times 27.98191] \times 931.5 MeV$$
$$Q= -0.02888 \times 931.5 MeV = -26.9 MeV$$
Since Q- value is negative, fission is not possible energetically.
13.7 The fission properties of ${_{94}\textrm{Pu}^{239}}$ are very similar to those of ${_{92}\textrm{U}^{235}}$ . 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 ${_{94}\textrm{Pu}^{239}}$ undergo fission?
Solution:
No of atoms in 239g of ${_{94}\textrm{Pu}^{239}}= 6.023 \times 10^{23}$
No of atoms in 1 kg (=1000g) of ${_{94}\textrm{Pu}^{239}}$
$$N= \frac{6.023 \times 10^{23}}{239} \times 1000 $$
$$N= 2.52 \times 10^{24}$$
Energy released/fission = 180MeV
Total energy released
$$T.E= 2.52 \times 10^{24} \times 180$$
$$T.E= 4.54 \times 10^{26}MeV$$
13.8 How long can an electric lamp of 100W be kept glowing by fusion of 2.0 kg of deuterium? Take the fusion reaction as
${_{1}\textrm{H}^{2}} + {_{1}\textrm{H}^{2}} \rightarrow {_{2}\textrm{H}^{3}} + n + 3.27 MeV$
Solution:
Number of atoms in 2g of deuterium = $6.023 \times 10^{23}$
Number of atoms in 2.0 kg (=2000g) of deuterium
$$N= \frac{6.023 \times 10^{23}}{2} \times 2000$$
$$N= 6.023 \times10^{26}$$
Energy released in the fusion of two deuterium nuclei = 3.27 MeV
Energy released in the fusion of one deuterium nuclei = $\frac{3.27}{2}$ MeV
Energy released in the fusion of one deuterium nuclei = 1.635 MeV
Total energy released in the fusion of 2.0 kg of deuterium
$$E= 1.635MeV \times 6.023 \times 10^{26}$$
$$E= 1.635 \times 10^{6} \times 1.6 \times 10^{-19} \times 6.023 \times 10^{26}$$
$$E= 1.635 \times 6.023 \times 10^{26} \times 1.6 \times 10^{-13}J$$
$$E= 15.75 \times 10^{13}J$$
power of bulb , P = 100 W = 100 J/sec
time for which the bulb will glow is
$$t = \frac{E}{P} = \frac{15.75 \times 10^{13}}{100}s$$
$$t = \frac{15.75 \times 10^{13}}{100 \times 3.15 \times 10^{7}}years$$
$$t = 4.99 \times 10^{4} years$$
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 fm.)
solution:
For head on collision, distance between the centres of two deuterons is
$$r = 2 \times Radius$$
$$r = 2 \times 2 fm$$
$$r = 4 fm$$
$$r = 4 \times 10^{-15}m$$
charge on deuteron , e = charge on proton
$$e = 1.6 \times 10^{-19}C$$
The initial mechanical energy E of the two deuterons before collision is
$E = 2 \times K.E$ of each deuteron
when the two deuterons stop , their total energy is entirely potential energy.
P.E of two deuterons during head-on collision is
$$P.E = \frac{1}{4 \pi \varepsilon_{0}}\frac{e^{2}}{r}$$
$$P.E = 9 \times 10^{9} \times \frac{(1.6 \times 10^{-19})^{2}}{4 \times 10^{-15}}J$$
$$P.E= 9 \times 10^{9} \times \frac{(1.6 \times 10^{-19})^{2}}{4 \times 10^{-15}} \times \frac{1}{1.6 \times 10^{-16}}keV$$
$$P.E = 360 keV$$
K.E of each deuteron = 360/2 = 180 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).
solution :
mass of nucleus = A a.m.u = $A \times 1.66 \times 10^{-27}kg$
Volume of nucleus
$$V = \frac{4}{3} \pi R^{3}$$
$$V = \frac{4}{3} \pi (R_{0}A^{1/3})^{3}$$
$$V = \frac{4}{3} \pi R_{0}^{3}A$$
Now , Nuclear density =
$$\rho = \frac{Mass \ of \ nucleus}{volume \ of \ nuleus}$$
$$\rho = \frac{A \times 1.66 \times 10^{-27}}{(4/3) \pi R_{0}^{3}A}$$
$$\rho = \frac{3 \times 1.66 \times 10^{-27}}{4 \pi R_{0}^{3}}$$
Therefore , nuclear matter density is independent of A.