Nuclear Fission, Fusion and Nuclear Reactor Questions in English

(C) The reaction is ${ }_1 H^2 + { }_1 H^2 \rightarrow { }_2 He^4 + Q$.
Total binding energy of reactants: Each deuteron has $2$ nucleons,and there are $2$ deuterons. Total nucleons = $4$. Binding energy per nucleon = $1.15 \text{ MeV}$. Total binding energy = $4 \times 1.15 \text{ MeV} = 4.6 \text{ MeV}$.
Total binding energy of products: The $\alpha$-particle $({ }_2 He^4)$ has $4$ nucleons. Binding energy per nucleon = $7.1 \text{ MeV}$. Total binding energy = $4 \times 7.1 \text{ MeV} = 28.4 \text{ MeV}$.
Energy released $(Q)$ = (Total binding energy of products) - (Total binding energy of reactants) = $28.4 \text{ MeV} - 4.6 \text{ MeV} = 23.8 \text{ MeV}$.
The energy released per nucleon is calculated by dividing the total energy released by the total number of nucleons in the product nucleus $({ }_2 He^4)$,which is $4$.
Energy released per nucleon = $23.8 \text{ MeV} / 4 = 5.95 \text{ MeV}$.

(D) Given: Power $P = 12 \text{ MW} = 12 \times 10^6 \text{ J/s}$.
Time $t = 1 \text{ day} = 24 \times 3600 \text{ s} = 86400 \text{ s}$.
Energy released per fission $E = 200 \text{ MeV} = 200 \times 10^6 \times 1.6 \times 10^{-19} \text{ J} = 3.2 \times 10^{-11} \text{ J}$.
Total energy produced in one day $E_{total} = P \times t = 12 \times 10^6 \times 86400 \text{ J} = 1.0368 \times 10^{12} \text{ J}$.
Number of fissions $n = \frac{E_{total}}{E} = \frac{1.0368 \times 10^{12}}{3.2 \times 10^{-11}} = 3.24 \times 10^{22}$.
Mass of one atom of ${}_{92}U^{235} = \frac{235}{6.023 \times 10^{23}} \text{ g}$.
Total mass consumed $m = n \times \text{mass of one atom} = \frac{3.24 \times 10^{22} \times 235}{6.023 \times 10^{23}} \approx 12.64 \text{ g}$.

(A) Energy released per fission per atom, $E = 200 \text{ MeV} = 200 \times 10^6 \times 1.6 \times 10^{-19} \text{ J} = 3.2 \times 10^{-11} \text{ J}$.
Number of atoms in $0.1 \text{ kg}$ of ${ }_{92}^{235} U$ is $N = \frac{m}{M} \times N_A = \frac{0.1 \text{ kg}}{235 \times 10^{-3} \text{ kg/mol}} \times 6.023 \times 10^{23} \text{ atoms/mol} \approx 2.563 \times 10^{23} \text{ atoms}$.
Total energy released in Joules is $E_{total} = N \times E = 2.563 \times 10^{23} \times 3.2 \times 10^{-11} \text{ J} \approx 8.2016 \times 10^{12} \text{ J}$.
Since $1 \text{ kWh} = 3.6 \times 10^6 \text{ J}$, the energy in $\text{kWh}$ is $E_{kWh} = \frac{8.2016 \times 10^{12}}{3.6 \times 10^6} \approx 2.278 \times 10^6 \text{ kWh} \approx 22.8 \times 10^5 \text{ kWh}$.

(D) Nuclear fuels are materials that can be consumed by nuclear fission or fusion to produce energy. Uranium,thorium,and plutonium are well-known radioactive elements used in nuclear reactors. Titanium is a transition metal known for its high strength-to-weight ratio and corrosion resistance,but it is not a radioactive material capable of sustaining a nuclear chain reaction. Therefore,it is not used as a nuclear fuel.

(B) Deep inside the core of stars,protons collide with each other at extremely high speeds due to high temperature and pressure. These collisions result in the formation of a helium nucleus,releasing a tremendous amount of energy in the process. This specific nuclear reaction is known as nuclear fusion.

(B) Nuclear fusion is a process in which two light nuclei combine to form a heavier nucleus. This process requires overcoming the strong electrostatic repulsion between the positively charged nuclei. To achieve this,the nuclei must have extremely high kinetic energy,which is provided by maintaining a very high temperature (on the order of $10^7$ to $10^8 \ K$). Therefore,fusion reactions are initiated with the help of high temperature.

(B) Given,energy released per nucleus fission,$E_0 = 188 MeV$ and mass,$m = 100 g$.
First,calculate the number of nuclei $N$ in $100 g$ of ${ }_{92} U^{235}$:
$N = \frac{m}{M} \times N_A = \frac{100}{235} \times 6.022 \times 10^{23} \approx 2.56 \times 10^{23}$ nuclei.
Next,convert the energy released per nucleus into Joules:
$E_0' = 188 \times 10^6 \times 1.6 \times 10^{-19} J = 300.8 \times 10^{-13} J$.
The total energy released $E$ for $N$ nuclei is:
$E = N \times E_0' = (2.56 \times 10^{23}) \times (300.8 \times 10^{-13} J) \approx 7.71 \times 10^{12} J$.
Thus,the correct option is $B$.

(D) According to the law of conservation of mass-energy,the total mass-energy before the reaction must equal the total mass-energy after the reaction.
Initial mass = $m + m = 2m$.
Final mass = $M_{He} + \frac{Q}{c^2}$ (where $M_{He}$ is the mass of the helium nucleus).
Equating the two: $2m = M_{He} + \frac{Q}{c^2}$.
Therefore,the mass of the helium nucleus is $M_{He} = 2m - \frac{Q}{c^2}$.

(A) To initiate a nuclear fusion reaction, the kinetic energy of the nuclei must be sufficient to overcome the repulsive electrostatic potential energy barrier.
According to the kinetic theory of gases, the average kinetic energy of a particle at temperature $T$ is given by $E = \frac{3}{2} kT$, where $k$ is the Boltzmann constant.
Equating the kinetic energy to the potential energy barrier $U = 2.07 \times 10^{-14} \,J$:
$\frac{3}{2} kT = U$
$T = \frac{2U}{3k}$
Substituting the given values:
$T = \frac{2 \times 2.07 \times 10^{-14}}{3 \times 1.38 \times 10^{-23}}$
$T = \frac{4.14 \times 10^{-14}}{4.14 \times 10^{-23}}$
$T = 10^9 \,K$
Therefore, the required temperature is $10^9 \,K$.

(B) Total energy required $E = P \times t = (3.70 \times 10^7 \text{ J/s}) \times (144 \times 10^4 \text{ s}) = 5.328 \times 10^{13} \text{ J}$.
Energy released per fission $E_f = 185 \text{ MeV} = 185 \times 10^6 \times 1.6 \times 10^{-19} \text{ J} = 2.96 \times 10^{-11} \text{ J}$.
Number of fissions $N = E / E_f = (5.328 \times 10^{13}) / (2.96 \times 10^{-11}) = 1.8 \times 10^{24} \text{ atoms}$.
Mass of fuel $m = (N / N_A) \times M = (1.8 \times 10^{24} / 6 \times 10^{23}) \times 235 \text{ g} = 0.3 \times 235 \text{ g} = 70.5 \text{ g} = 0.0705 \text{ kg}$.
Re-evaluating the calculation: $N = 1.8 \times 10^{24}$ atoms. Mass $= (1.8 \times 10^{24} / 6 \times 10^{23}) \times 235 = 3 \times 235 = 705 \text{ g} = 0.705 \text{ kg}$.

(B) The rate of mass consumption is given by $\frac{\Delta m}{\Delta t} = 1 \times 10^{-3} \text{ g s}^{-1} = 10^{-6} \text{ kg s}^{-1}$.
According to Einstein's mass-energy equivalence principle,the power $P$ generated is given by $P = \frac{\Delta E}{\Delta t} = \left(\frac{\Delta m}{\Delta t}\right) c^2$.
Substituting the values,where $c = 3 \times 10^8 \text{ m s}^{-1}$:
$P = 10^{-6} \text{ kg s}^{-1} \times (3 \times 10^8 \text{ m s}^{-1})^2$.
$P = 10^{-6} \times 9 \times 10^{16} \text{ W}$.
$P = 9 \times 10^{10} \text{ W}$.
Since $1 \text{ kW} = 10^3 \text{ W}$,we convert the power to kW:
$P = \frac{9 \times 10^{10}}{10^3} \text{ kW} = 9 \times 10^7 \text{ kW}$.

(A) In the Sun,the primary source of energy is the proton-proton cycle.
In this cycle,hydrogen nuclei undergo a series of nuclear fusion reactions to form helium nuclei.
During this fusion process,a significant amount of mass is converted into energy according to Einstein's mass-energy equivalence principle,$E = mc^2$.

(A-R, B-P, C-Q, D-S) The correct matches are as follows:
$A$. Rocket propulsion is based on Newton's third law of motion,which corresponds to $R$.
$B$. The lift of an aeroplane is explained by Bernoulli's principle in fluid dynamics,which corresponds to $P$.
$C$. Optical fibres work on the principle of total internal reflection of light,which corresponds to $Q$.
$D$. Fusion test reactors use magnetic confinement of plasma to sustain the fusion process,which corresponds to $S$.
Therefore,the correct matching is $A-R, B-P, C-Q, D-S$.

(C) Magnetic confinement fusion is an approach to generate thermonuclear fusion power that uses magnetic fields to confine fusion fuel in the form of a plasma.
In order to overcome the electrostatic repulsion between the nuclei,they must have a temperature of tens of millions of degrees,creating a plasma.
This plasma is then confined using strong magnetic fields to prevent it from touching the reactor walls.

(C) The scientific principle that forms the basis of the Tokamak technology is the magnetic confinement of plasma.
In a Tokamak,a strong magnetic field is used to confine hot plasma in the shape of a torus (a doughnut shape).
This magnetic confinement prevents the high-temperature plasma from touching the walls of the reactor,which is essential for achieving controlled thermonuclear fusion.

(B) In the fission of ${ }_{92}^{235} U$,the neutrons produced are fast neutrons.
These neutrons have a kinetic energy distribution,but their average energy is approximately $2 \,MeV$.
We know that $1 \,eV = 1.6 \times 10^{-19} \,J$.
Therefore,$2 \,MeV = 2 \times 10^6 \times 1.6 \times 10^{-19} \,J$.
$2 \,MeV = 3.2 \times 10^{-13} \,J$.
This can be written as $320 \times 10^{-15} \,J$.

(C) Power $P = 64 \text{ kW} = 64 \times 10^3 \text{ W} = 64 \times 10^3 \text{ J/s}$.
Total energy produced in one hour $(t = 3600 \text{ s})$ is $E = P \times t = 64 \times 10^3 \times 3600 \text{ J} = 2304 \times 10^5 \text{ J} = 2.304 \times 10^8 \text{ J}$.
Number of nuclei $N = 7.2 \times 10^{18}$.
Energy released per fission $\epsilon = E / N = (2.304 \times 10^8) / (7.2 \times 10^{18}) \text{ J}$.
$\epsilon = 0.32 \times 10^{-10} \text{ J}$.

(D) In the decay of a nucleus with mass number $A = 236$,the $\alpha$-particle $(m_{\alpha} = 4)$ and the daughter nucleus $(m_d = 232)$ are produced.
Given the kinetic energy of the $\alpha$-particle is $(KE)_{\alpha} = E$.
Since the total momentum is conserved,the magnitudes of the momenta of the $\alpha$-particle and the daughter nucleus are equal: $P_{\alpha} = P_d = P$.
The kinetic energy is given by $KE = \frac{P^2}{2m}$,which implies $KE \propto \frac{1}{m}$.
Therefore,the ratio of kinetic energies is $\frac{(KE)_d}{(KE)_{\alpha}} = \frac{m_{\alpha}}{m_d} = \frac{4}{232} = \frac{1}{58}$.
Thus,$(KE)_d = \frac{E}{58}$.
The total energy released $(Q)$ is the sum of the kinetic energies of the products: $Q = (KE)_{\alpha} + (KE)_d = E + \frac{E}{58} = \frac{59 E}{58}$.

(C) The number of moles $n$ is given by $n = \frac{m}{M} = \frac{N}{N_A}$,where $m$ is the mass,$M$ is the molar mass,$N$ is the number of atoms,and $N_A$ is Avogadro's number.
Given $m = 1 \text{ mg} = 10^{-3} \text{ g}$,$M = 240 \text{ g/mol}$,and $N_A = 6 \times 10^{23} \text{ atoms/mol}$.
The number of atoms $N$ is:
$N = \frac{m}{M} \times N_A = \frac{10^{-3}}{240} \times 6 \times 10^{23} = 2.5 \times 10^{18} \text{ atoms}$.
The energy released per fission is $E_f = 200 \text{ MeV} = 200 \times 10^6 \times 1.6 \times 10^{-19} \text{ J} = 3.2 \times 10^{-11} \text{ J}$.
Total energy released $E = N \times E_f = 2.5 \times 10^{18} \times 3.2 \times 10^{-11} \text{ J} = 8.0 \times 10^7 \text{ J}$.

(D) Energy released per fission $E = 200 \text{ MeV} = 200 \times 10^6 \times 1.6 \times 10^{-19} \text{ J} = 3.2 \times 10^{-11} \text{ J}$.
Power $P = 128 \text{ W} = 128 \text{ J/s}$.
Let $n$ be the number of fissions per second.
Then,$P = n \times E$.
$n = \frac{P}{E} = \frac{128}{3.2 \times 10^{-11}}$.
$n = \frac{1280}{3.2} \times 10^{10} = 400 \times 10^{10} = 4 \times 10^{12} \text{ fissions/s}$.

(C) The neutron multiplication factor $K$ (also denoted as $k$) is defined as the ratio of the number of neutrons produced in a given generation to the number of neutrons produced in the preceding generation.
When $K = 1$,the rate of neutron production is equal to the rate of neutron loss,meaning the chain reaction is self-sustaining at a constant power level.
This state is known as the critical state of a nuclear reactor.
If $K < 1$,the reactor is subcritical,and the chain reaction dies out.
If $K > 1$,the reactor is supercritical,and the power level increases exponentially.

(C) In nuclear fission and fusion, energy is released due to the change in mass of the nuclei.
According to Einstein's mass-energy equivalence principle, mass can be converted into energy and vice-versa, given by the equation $E = mc^2$.
In fission, a heavy nucleus splits into lighter nuclei, and in fusion, lighter nuclei combine to form a heavier nucleus.
In both processes, the total mass of the products is slightly less than the total mass of the reactants.
This mass defect $(\Delta m)$ is converted into energy according to the equation $E = (\Delta m)c^2$.
Therefore, both nuclear fission and fusion are explained by Einstein's mass-energy equation.

(C) In a nuclear reaction,both the mass number and the atomic number are conserved.
Given reaction: ${ }_{13} Al^{27} + { }_2 He^4 \rightarrow { }_0 n^1 + X$
Conservation of mass number $(A)$:
$27 + 4 = 1 + A_X$
$31 = 1 + A_X \Rightarrow A_X = 30$
Conservation of atomic number $(Z)$:
$13 + 2 = 0 + Z_X$
$15 = Z_X$
Since the atomic number is $15$,the element is Phosphorus $(P)$.
Therefore,$X = { }_{15} P^{30}$.
Hence,option $C$ is correct.

(D) The number of atoms in $235 \text{ g}$ of $U^{235}$ is equal to the Avogadro number,$N_A = 6.02 \times 10^{23}$.
Energy released per nucleus $= 188 \text{ MeV} = 188 \times 1.6 \times 10^{-13} \text{ J} = 3.008 \times 10^{-11} \text{ J}$.
Total energy released $= N_A \times \text{Energy per nucleus}$.
Total energy $= (6.02 \times 10^{23}) \times (3.008 \times 10^{-11} \text{ J}) \approx 18.11 \times 10^{12} \text{ J}$.

(D) In a nuclear reactor,fast-moving neutrons produced during fission have high kinetic energy.
To sustain a chain reaction,these neutrons need to be slowed down to thermal energies so they can effectively cause further fission in $U^{235}$ nuclei.
Heavy water $(D_2O)$ is used as a moderator because it is effective at slowing down these fast-moving neutrons through elastic collisions without absorbing them significantly.

(A) Power $P = 5 \text{ mW} = 5 \times 10^{-3} \text{ W}$.
Energy released per fission $E = 200 \text{ MeV} = 200 \times 10^6 \times 1.6 \times 10^{-19} \text{ J} = 3.2 \times 10^{-11} \text{ J}$.
Let $n$ be the number of fissions per second.
Power $P = n \times E$.
Therefore,$n = \frac{P}{E} = \frac{5 \times 10^{-3} \text{ J/s}}{3.2 \times 10^{-11} \text{ J/fission}}$.
$n = \frac{5}{3.2} \times 10^8 = 1.5625 \times 10^8 \text{ fissions/s}$.
Thus,the number of fissions per second is $1.56 \times 10^8$.

(B) tokamak reactor uses magnetic fields to confine hot plasma in a doughnut-shaped region.
Plasma is heated to extremely high temperatures to achieve conditions suitable for thermonuclear fusion.
The primary mechanism for maintaining the stability and position of this high-temperature plasma is magnetic confinement.

(B) Total energy required $E = \text{Power} \times \text{Time} = (3.70 \times 10^7 \text{ J/s}) \times (144 \times 10^4 \text{ s}) = 5.328 \times 10^{13} \text{ J}$.
Energy per fission $\epsilon = 185 \text{ MeV} = 185 \times 10^6 \times 1.6 \times 10^{-19} \text{ J} = 2.96 \times 10^{-11} \text{ J}$.
Number of fissions required $N = \frac{E}{\epsilon} = \frac{5.328 \times 10^{13}}{2.96 \times 10^{-11}} = 1.8 \times 10^{24} \text{ atoms}$.
Mass of fuel $m = \frac{N \times M}{N_A} = \frac{1.8 \times 10^{24} \times 235}{6 \times 10^{23}} = 705 \text{ g} = 0.705 \text{ kg}$.

(A) Energy released in the fission of one nucleus is $E_1 = 200 \text{ MeV}$.
Converting this energy into Joules:
$E_1 = 200 \times 1.6 \times 10^{-13} \text{ J} = 3.2 \times 10^{-11} \text{ J}$.
We need to find the number of nuclei $(n)$ required to release a total energy $E_{total} = 1000 \text{ J}$.
The relationship is $E_{total} = n \times E_1$.
Therefore,$n = \frac{E_{total}}{E_1} = \frac{1000}{3.2 \times 10^{-11}}$.
$n = \frac{10^3}{3.2 \times 10^{-11}} = \frac{1}{3.2} \times 10^{14} = 0.3125 \times 10^{14} = 3.125 \times 10^{13}$ nuclei.

(C) The nuclear force acting between proton-neutron $(p-n)$,proton-proton $(p-p)$,and neutron-neutron $(n-n)$ are approximately equal and charge independent. Therefore,statement $A$ is wrong.
In a nuclear reactor,the neutron reproduction factor (often denoted as $k$) represents the ratio of the number of neutrons produced in one generation to the number of neutrons in the preceding generation. If $k > 1$,the chain reaction is supercritical and the fission rate increases exponentially (accelerating state). Therefore,statement $B$ is correct.

(B) In a nuclear reactor,the neutrons produced during fission are fast-moving. These fast neutrons are less likely to cause further fission in $U^{235}$ nuclei. The moderator (such as heavy water,graphite,or ordinary water) is used to slow down these fast neutrons to thermal energies,making them more effective at sustaining the chain reaction. Control rods are used to absorb excess neutrons to regulate the reaction rate,while coolants are used to remove excess heat.

(A) The number of uranium atoms $N$ in $5 \ mg$ of ${}^{235}U$ is given by $N = \frac{m}{M} \times N_A$,where $m = 5 \times 10^{-3} \ g$,$M = 235 \ g/mol$,and $N_A = 6.022 \times 10^{23} \ atoms/mol$.
$N = \frac{5 \times 10^{-3}}{235} \times 6.022 \times 10^{23} \approx 1.28 \times 10^{19} \ atoms$.
The total energy released $E$ is $N \times E_{fission}$,where $E_{fission} = 200 \ MeV = 200 \times 10^6 \times 1.6 \times 10^{-19} \ J = 3.2 \times 10^{-11} \ J$.
$E = 1.28 \times 10^{19} \times 3.2 \times 10^{-11} \ J \approx 4.096 \times 10^8 \ J$.
Thus,the approximate energy released is $4 \times 10^8 \ J$.

(A) The binding energy per nucleon $(BE/A)$ for a nucleus of mass number $A = 238$ is $7.6 \text{ MeV}$, and for a nucleus of mass number $A = 119$, it is $8.6 \text{ MeV}$.
Total binding energy of the parent nucleus $(A = 238)$ is $E_1 = 238 \times 7.6 \text{ MeV} = 1808.8 \text{ MeV}$.
When the nucleus splits into two fragments of mass number $119$ each, the total binding energy of the product nuclei is $E_2 = 2 \times (119 \times 8.6 \text{ MeV}) = 238 \times 8.6 \text{ MeV} = 2046.8 \text{ MeV}$.
The energy released in the fission process is $\Delta E = E_2 - E_1 = 2046.8 \text{ MeV} - 1808.8 \text{ MeV} = 238 \text{ MeV}$.
Among the given options, $214 \text{ MeV}$ is the closest approximate value.

(D) The energy released per fission is $E_{fission} = 200 MeV$.
Converting this energy into Joules: $E_{fission} = 200 \times 10^6 \times 1.6 \times 10^{-19} J = 3.2 \times 10^{-11} J$.
The power required is $P = 3.2 W$,which means $3.2 J$ of energy is required per second.
The number of fissions per second $(n)$ is given by the ratio of total power to energy per fission:
$n = \frac{P}{E_{fission}} = \frac{3.2 J/s}{3.2 \times 10^{-11} J} = 10^{11} \text{ fissions/second}$.

(B) The molar mass of $_{92}^{235} U$ is $235 \text{ g/mol}$.
Number of moles in $47 \text{ g}$ of $_{92}^{235} U$ is $n = \frac{47 \text{ g}}{235 \text{ g/mol}} = 0.2 \text{ moles} = \frac{1}{5} \text{ moles}$.
Total number of atoms $N = n \times N_A = \frac{1}{5} \times 6 \times 10^{23} = 1.2 \times 10^{23} \text{ atoms}$.
Energy released per fission is $190 \text{ MeV}$.
Total energy released $= N \times 190 \text{ MeV} = (1.2 \times 10^{23}) \times 190 \text{ MeV} = 228 \times 10^{23} \text{ MeV}$.
Comparing this with $\alpha \times 10^{23} \text{ MeV}$,we get $\alpha = 228$.

(C) The mass defect $\Delta m$ is the difference between the mass of the products and the initial mass.
$\Delta m = (4 \times 4.002 \ amu) - 15.348 \ amu = 16.008 \ amu - 15.348 \ amu = 0.66 \ amu$.
Convert the mass defect to $kg$: $\Delta m = 0.66 \times 1.66 \times 10^{-27} \ kg = 1.0956 \times 10^{-27} \ kg$.
The energy required $E = \Delta m c^2 = 1.0956 \times 10^{-27} \times (3 \times 10^8)^2 = 1.0956 \times 10^{-27} \times 9 \times 10^{16} = 9.8604 \times 10^{-11} \ J$.
Using $E = h\nu$,the frequency $\nu = E / h = (9.8604 \times 10^{-11}) / (6.6 \times 10^{-34}) \approx 1.494 \times 10^{23} \ Hz$.
Converting to $kHz$: $\nu = 1.494 \times 10^{23} / 10^3 = 1.494 \times 10^{20} \ kHz$.

(B) The proton-proton chain reaction,which is the primary fusion process in the Sun,can be represented by the net equation:
$4_1^1H + 2e^- \rightarrow ^4_2He + 2\nu_e + 26.7 \text{ MeV}$.
In this process,four hydrogen nuclei (protons) fuse to form one helium nucleus $(^4_2He)$,releasing two neutrinos $(
u_e)$ and a total energy of approximately $26.7 \text{ MeV}$.

(B) The nuclear fusion reaction is: $2 \times ^{2}_{1}H \to ^{4}_{2}He$.
Binding energy of one $^{2}_{1}H$ nucleus $= 2 \times 1.1 \text{ MeV} = 2.2 \text{ MeV}$.
Total binding energy of reactants $= 2 \times 2.2 \text{ MeV} = 4.4 \text{ MeV}$.
Binding energy of one $^{4}_{2}He$ nucleus $= 4 \times 7.2 \text{ MeV} = 28.8 \text{ MeV}$.
Energy released $= \text{Binding energy of product} - \text{Binding energy of reactants}$.
Energy released $= 28.8 \text{ MeV} - 4.4 \text{ MeV} = 24.4 \text{ MeV}$.

(B) The nuclear reaction is: $^7_3\text{Li} + ^1_1\text{H} \rightarrow 2(^4_2\text{He})$.
Mass of reactants $= 7.0183 \text{ u} + 1.008 \text{ u} = 8.0263 \text{ u}$.
Mass of products $= 2 \times 4.004 \text{ u} = 8.008 \text{ u}$.
Mass defect $\Delta m = 8.0263 \text{ u} - 8.008 \text{ u} = 0.0183 \text{ u}$.
Energy released per reaction $= 0.0183 \times 931 \text{ MeV} = 17.0373 \text{ MeV} = 1.70373 \times 10^7 \text{ eV}$.
Number of moles in $\frac{7}{17.13} \text{ kg}$ (i.e.,$\frac{7000}{17.13} \text{ g}$) of $^7_3\text{Li}$ (molar mass $\approx 7 \text{ g/mol}$): $n = \frac{7000 / 17.13}{7} = \frac{1000}{17.13} \approx 58.377 \text{ mol}$.
Total number of atoms $N = n \times N_A = 58.377 \times 6.0 \times 10^{23} \approx 3.5026 \times 10^{25}$.
Total energy released $= N \times (1.70373 \times 10^7 \text{ eV}) \approx 3.5026 \times 10^{25} \times 1.70373 \times 10^7 \text{ eV} \approx 5.967 \times 10^{32} \text{ eV}$.
Comparing with $\alpha \times 10^{32} \text{ eV}$,we get $\alpha \approx 5.967$. The nearest integer is $6$.