5 Marks Questions

Question 15 Marks

How long can an electric lamp of 100 W be kept glowing by fusion of 2.0 kg of deuterium? Take the fusion reaction as :
${ }_1^2 H +{ }_1^2 H \rightarrow{ }_2^3 He +n+3.27 MeV$

Answer

Amount of energy released in the fusion of 2 atoms of deuterium = 3.2 MeV.
Amount of deuterium in use = 2 kg
Number of deuterium atoms or nuclei in 2 kg of ${ }_1 H ^2$ $($deuterium$)=6.023 \times 10^{23}$
Number of nuclei or deuterium atoms in 2 kg of ${ }_1 H ^2$ or 2000 g.
$=\frac{6.023 \times 10^{23}}{2} \times 2000$
$=6.023 \times 10^{26}$
Energy released in the fusion of 2 nuclei
= 3.2 MeV
Energy released in the fusion of $6.023 \times 10^{26}$ nuclei
$=\frac{3.2}{2} \times 6.023 \times 10^{26}$
$\begin{array}{l}=9.6368 \times 10^{26} MeV \\ =9.6368 \times 10^{26} \times 1.6 \times 10^{-13} J \\ =15.42 \times 10^{13} J \\ =15.42 \times 10^{13} WS \end{array}$
Power $P =100 W$
Consider this energy can be kept illuminated for $t$ seconds.
Hence the above mentioned electric power
$=100 t$
$100 t=15.42 \times 10^{13}$
or$\quad$$t=\frac{15.42 \times 10^{13}}{100}$
$=15.42 \times 10^{11}$ seconds
$=\frac{15.42 \times 10^{11}}{365 \times 60 \times 60 \times 24}$ years
or$\quad$$t=4.9 \times 10^4$ years

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Question 25 Marks

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 made entirely of ${ }_{29}^{63} Cu$ atoms (of mass 62.92960 u).

Answer

Mass of atom $=62.92960 u$
$ \text { Mass of } 29 \text { electrons } =29 \times 0.000548 u $
$=0.015892 u $
$ \text { Mass of nucleus } =(62.92960-0.015892) u $
$=62.913708 u $
$ \text { Mass of } 29 \text { protons } =29 \times 1.007825 u $
$=29.226925 u $
Mass of $(63-29=34)$ neutrons
$=34 \times 1.008665 u $
$=34.29461 u $
Total mass of proton and neutron
$=(29.226925+34.29461) u $
$=63.521535 u$
$\text {Binding energy }=(63.521535-62.913708) \times 931.5 MeV $
$=0.607827 \times 931.5 MeV$
Amount of binding energy for 1 copper nucleus
$=0.607827 \times 931.5 MeV$
In 63 g copper, number of copper atoms = Avogadro number
$N =6.023 \times 10^{23}$
$\therefore \quad$ In 1 g copper $=\frac{6.023 \times 10^{23}}{63}$
Hence, in 3 g copper $=\frac{6.023 \times 10^{23}}{63} \times 3$
Hence the binding energy required will be
$= \frac{6.023 \times 10^{23}}{63} \times 3 $
$\times 0.607827 \times 931.5 MeV $
$\simeq 1.584 \times 10^{25} MeV$
$=1.584 \times 10^{25} \times 10^6 \times 1.6 \times 10^{-19} J$
$\simeq 2.535 \times 10^{12} J$

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Question 35 Marks

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

Answer

(i) ${ }_{26}^{56} Fe$ has 26 protons and 56 - 26 = 30 neutrons.
$\begin{aligned} \text { Mass of } 26 \text { protons } =26 \times 1.007825 \\ =26.20345 \text { a.m.u. }\end{aligned}$
$\begin{aligned} \text { Mass of } 30 \text { neutrons } =30 \times 1.008665 \\ =30.25995 \text { a.m.u. }\end{aligned}$
Total mass of 25 nucleons
$\begin{array}{l}=26.20345+30.25995 \\ =56.46340 \text { a.m.u. }\end{array}$
Mass of ${ }_{26}^{56} Fe$ nucleus $=55.934939$ a.m.u.
Hence mass loss $\Delta m=56.46340-55.934939$
$=0.528461$ a.m.u.
$\therefore$ Total binding energy $=0.528461 \times 931.5 MeV$
$=492.26 MeV$
Antinucleon mean binding energy
$\overline{ B }=\frac{\Delta E }{ A }=\frac{492.26}{56}=8.790 MeV$
(ii) ${ }_{83}^{209} Bi$ nucleus contains 83 protons and $(209-83)$ $=126$ atoms.
$\begin{aligned} \text { Mass of } 83 \text { protons } =83 \times 1.007825 \\ =83.64975 \text { a.m.u. }\end{aligned}$
$\begin{aligned} \text { Mass of } 126 \text { neutrons } =126 \times 1.008665 \\ =127.09190 \text { a.m.u. }\end{aligned}$
Total mass of nucleon $=210.741260$ a.m.u.
Mass of ${ }_{83}^{209} Bi$ nucleus $=208.980388$ a.m.u.
Therefore mass defect
$\begin{array}{l} \Delta m=210.741260 \text { a.m.u. } -208.980388 \text { a.m.u. }\end{array}$
$=1.760872$ a.m.u.
Total binding energy $=1.760872 \times 931.5 MeV$
$=1640.26 MeV$
Mean binding energy per nucleon
$\begin{aligned} \overrightarrow{ B } =\frac{\Delta E }{ A }=\frac{1640.26}{209} MeV \\ =7.84 MeV \end{aligned}$
Therefore, the binding energy per nucleon of ${ }_{26}^{56} Fe$ is greater than that of ${ }_{83}^{209} Bi$.

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