Physics M.C.Q (1 Marks)

$W = {E_f} - {E_i} = 0 - {E_i} = - \left( { - \frac{{13.6}}{{{n^2}}}eV} \right)$

$ \Rightarrow W = \frac{{13.6}}{{{{(2)}^2}}} \times 1.6 \times {10^{ - 19}}J$

$= 3.4 \times 1.6 \times 10^{-19}J.$

Total energy of $A=110 \times 8.2 \mathrm{MeV}$

Total energy of $B=90 \times 78.2 \mathrm{MMeV}$

So, $2 \omega \times 7.4=(110 \times 8.2)+(90 \times 8.2)$

Hence - energy $=-2 \omega \times 7.4+(110 \times 8.2)+(90 \times 8.2)$

$=160 \mathrm{MeV}$

Energy required to remove one neutron = Difference in total binding energy $=13 \times 7.5-12 \times 7.68 \mathrm{MeV}$

$=5.34 M e V$

Final energy $=E_2$

Initial energy $=2 E _1$ (two atoms of $X$ )

Hence, difference in binding energies is $=E_2-2 E_1$

Given that $Q$ is released.

Hence, $E _2-2 E _1= Q$

$_{1} H^{2}=n+p$

Total energy of deutron $=1476 \mathrm{MeV}$ Energy of neutron $+$ proton $=940+939=1879 \mathrm{MeV}$

Energy required to disintegrate Deutron $=3 M e V$

Qvalue $Q=50$

we know that,

The kinetic energy is

$E=\frac{A-4}{A} \times Q \ldots \ldots(I)$

Now, put the value in equation $(I)$

$48 A=(A-4) 50$

$48 A-50 A=-200$

$2 A=200$

$A=100$

Hence, The mass number of the mother nucleus is $100$

$B.E.$ $1692 \quad 1902.4$

energy of two $\gamma$ -rays photons $=2 \times 5.2=10.4 \mathrm{\,MeV}$

$\therefore$ usable energy per fission

${=(210.4-10.4) \mathrm{\,MeV}} $

${=200 \mathrm{\,MeV}}$

$=32 \mathrm{\,MW}=32 \times 10^{6} \mathrm{\,W}$

Hence number of fission of Uranium nuclei per second

$=\frac{32 \times 10^{6}}{200 \times 1.6 \times 10^{-13}}=1 \times 10^{18}$

$\frac{m}{4} v_{1}=\frac{3 m}{4} v_{2}$

$\Delta \mathrm{m} \mathrm{C}^{2}=\frac{1}{2} \frac{\mathrm{m}}{4} \mathrm{v}_{1}^{2}+\frac{1}{2} \times \frac{3 \mathrm{m}}{4} \mathrm{v}_{2}^{2}$

$\mathrm{v}_{1}=\sqrt{\frac{6 \Delta \mathrm{m}}{\mathrm{M}}}$

$(A+1) B E-8 A=3$

$\mathrm{BE}=\frac{8 \mathrm{A}+3}{\mathrm{A}+1}$

$\mathrm{KE}_{\alpha-\text { particle }}=(6.6)\left(\frac{216}{200}\right)=6.48 \mathrm{\,MeV}$

$\mathrm{Q}=\left(3 \times \mathrm{M}_{\mathrm{He}}-\mathrm{M}_{\mathrm{C}}\right) \times \mathrm{c}^{2}$

$\mathrm{Q}=7.27 \mathrm{\,MeV}$

As initial momentum was zero so final should also be zero.

so $p_{1}+p_{2}=0$ or $p_{1}=-p_{2}$ i.e. both have same values$........(1)$

where $p_{1}$ is momentum of $m_{1}$ and $p_{2}$ is that of $m_{2}$

The deBroglie wavelength is given by $\lambda=\frac{h}{p}$

As both mass have same momentum, $, p_{1}=-p_{2}$

so same values of wavelength. So answer is $\lambda$

$=2.224 \mathrm{\,MeV}$

Energy equivalent to $_{2} \mathrm{He}^{4}=4 \times 7.047$ $=28.188 \mathrm{\,MeV}$

From the equation, energy released $=28.188-2 \times 2.224=23.74 \mathrm{\,MeV} \approx 24 \mathrm{\,MeV}$

Given  $W = 2Y$

$BE$ of reactants $= 120 \times 7.5 = 900\, MeV$
$BE$ of products $= 2 \times (60 \times 8.5) = 1020\, MeV$

Initial $BE =7.4 \times 200 \,MeV$

Final BE $=8.2 \times 110+8.2 \times 90 \,MeV$

$\therefore$ Energy release $=$ Final $-$ Initial

$=(8.2-7.4) \times 200=0.8 \times 200$

$=160 \,MeV$

$KE$ of nuclei $=\frac{3}{2} k t=7.659 \times 10^{-14}$

(for fusion)

Potential energy between nuclei must be much less than the initiating $KE$ so that the nuclei have enough $KE$ for reaction to take place.

$_{92}{U^{236}}{ \to _{56}}B{a^{144}}{ + _{36}}K{r^{89}} + {3_0}{n^1} + Q$.

$ = 200 \times {10^6} \times 1.6 \times {10^{ - 19}}J = 3.2 \times {10^{ - 11}}J$

$P = 16\,KW = 16 \times {10^3}Watt$ 

Now, number of nuclei required per second 

$n = \frac{P}{E} = \frac{{16 \times {{10}^3}}}{{3.2 \times {{10}^{ - 11}}}} = 5 \times {10^{14}}$.

$ = \frac{{Power\;output}}{{Energy\;released\;per\;fission}}$

$ = \frac{{3.2 \times {{10}^6}}}{{200 \times {{10}^6} \times 1.6 \times {{10}^{ - 19}}}} = 1 \times {10^{17}}$ 

$ \Rightarrow $ Number of fission per minute $ = 60 \times {10^{17}} = 6 \times {10^{18}}$

$ = (2.0141) \times 2 = 4.0282\;amu$

Total mass of products $ = 4.0024\;amu$

Mass defect $ = 4.0282\;amu - 4.0024\;amu$

$ = 0.0258\;amu$

$\therefore $Energy released$E = 931 \times 0.0258 = 24\;MeV$

$ \Rightarrow 300 \times {10^6}$$ = \frac{{n \times 170 \times {{10}^6} \times 1.6 \times {{10}^{ - 19}}}}{t}$

Number of atoms per sec $\frac{n}{t} = 1.102 \times {10^{19}}$

Number of atoms per hour $= 1.02 × 1019 ×3600$

$= 3.97 × 10^{22}.$

Applying charge and mass conservation

$4 + A = A + 3 + x \Rightarrow x = 1$

==> $2 + z = z + 2 + n \Rightarrow n = 0$ 

Hence $A$ is a neutron.

${ }_{92} U ^{238} \rightarrow{ }_{90} Th ^{234}+{ }_2 He ^4$

Using conservation of linear moments

$206\,v' + 4v = 0 \Rightarrow v' = - \frac{{4v}}{{(206)}}$

$ \Rightarrow \,|v'|\, = \frac{{4v}}{{206}}$

$\xrightarrow{\alpha }{{\,}_{69}}{{A}_{3}}^{172}  \xrightarrow{\gamma }{{\,}_{69}}{{A}_{4}}^{172}$

and ${n_\beta } = (2{n_\alpha } - Z + Z') = (2 \times 6 - 90 + 82) = 4$

==> $A' = A - 4{n_\alpha } = 236 - 4 \times 3 = 224$

and $Z' = ({n_\beta } - 2{n_\alpha } + Z) = (1 - 2 \times 3 + 88) = 83$

This decreases atomic number to $90 - 4 \times 2 = 82$

Since atomic number of $_{83}{Y^{222}}$ is $83,$

this is possible if one $\beta - $ particle is emitted.

$_{82}{{Y}^{198}}{{\xrightarrow{\beta -\text{decay}}}_{83}}{{Z}^{198}}{{+}_{-1}}{{\beta }^{0}}$

${n_\beta } = 2{n_\alpha } - Z + Z' = 2 \times 6 - 90 + 82 = 4$

${n_\beta } = (2{n_\alpha } - Z + Z') = (2 \times 7 - 92 + 82) = 4$

So $ Z$ will remain same and $N$ will become $N - 4.$

Hence, $A = 234, Z = 91$

$ = 232 - 4 \times 6 = 208$ 

atomic number$Z' = Z + {n_\beta } - 2{n_\alpha }$

$ = 90 + 4 - 2 \times 6 = 82$

$X ={3_0}{n^1}$

Number of protons $= Z - 1,$

Number of neutrons $= (A - 12) - (Z - 1) = A - Z- 11$

$\therefore \frac{N}{P}  = \frac{{{\text{A - Z - 11}}}}{{{\text{Z - 1}}}}$

Number of protons $= (Z = 13)$

Number of neutrons $= (A -36) -(Z -13) = (A -Z -23)$

$\therefore$ $\frac{P}{N} = \frac{{(Z - 13)}}{{(A - Z - 23)}}$