Physics M.C.Q (1 Marks)

$A \rightarrow 286$

$ Z=81$

Left fraction of activity $\frac{1}{16}$

$\because \frac{ R }{ R _0} =\left(\frac{1}{2}\right)^{ t / T }$

$\frac{1}{16} =\left(\frac{1}{2}\right)^{ t / 20}$

$\left(\frac{1}{2}\right)^4 =\left(\frac{1}{2}\right)^{ t / 20}$

$4 =\frac{ t }{20}$

$t =80\,min$

$R=R_{0}(A)^{1 / 3}$

$\frac{R(125)}{R(64)}=\frac{R_{0}(125)^{1 / 3}}{R_{0}(64)^{1 / 3}}=\frac{5}{4}$

$T _{1 / 2}=$ time in which active no. of nuclei becomes half therefore statement $II$ is wrong.

This is $\beta^{+}$- decay

${ }_{11}^{22} Na \longrightarrow{ }_{10}^{22} Ne + e ^{+}+ v$

given binding energy per nucleon of $X, Y \& Z$ are

$7.6\, \mathrm{MeV}, 8.5\, \mathrm{MeV} \,\&\, 8.5\, \mathrm{MeV}$ respectively.

Gain in binding energy is:

$\mathrm{Q}=$ Binding Energy of products $-$ Binding energy of reactants

$=(120 \times 8.5 \times 2)-(240 \times 7.6)\, \mathrm{MeV}$

$=216\, \mathrm{MeV}$

$\alpha$ decreases atomic number by $2$

$\beta^{-}$increases atomic number by $1$

$A=-\frac{d N}{d t}=N \lambda$

$A = N \times \frac{0.693}{ T } \quad[ T$ is half life $]$

$A=2 \times 10^{21} \times \frac{0.693}{1.4 \times 10^{17}}$

$A=10^{4} B q$

$=0.5 \times 10^{-3} \times 9 \times 10^{16}$

$=4.5 \times 10^{13} J$

Hence due to gamma emission, there is no change in mass number and atomic number.

$2.2 \times 10^9=\frac{0.693}{\lambda}$

$\lambda=\frac{0.693}{2.2 \times 10^9}=3.15 \times 10^{-10}$

$R =\lambda N$

$N =\frac{ R }{\lambda}=\frac{10^{10}}{3.15 \times 10^{-10}}=3.17 \times 10^{19}$

$2$ protons and $2$ neutrons

$N=600-450=150$

According to the law of radioactive decay,

$\frac{N}{N_{0}}=\left(\frac{1}{2}\right)^{\frac{t}{T_{1 / 2}}} ;$ where $N_{0}$ is the number of nuclei initially.

$\therefore \quad \frac{150}{600}=\left(\frac{1}{2}\right)^{\frac{t}{T_{1 / 2}}} ;$ where $T_{1 / 2}=$ half life.

or $\quad\left(\frac{1}{2}\right)^{2}=\left(\frac{1}{2}\right)^{\frac{t}{T_{1 / 2}}}$

$\Rightarrow t=2 T_{10}=2 \times 10$ minutes $=20$ minutes

Then for $10^{20}$ fisssion/sec

$P=10^{20} \times 3 \cdot 2 \times 10^{-11} J / s$

$P=3.2 \times 10^{9} W$

$P=32 \times 10^{8} W$

$N(t)=N_{0} e^{-\lambda t}$

where $N_{0}$ is number of radioactive nuclei in the sample at some aribitrary time $t=0$ and $\lambda$ is the radioactive decay constant.

Given: $\lambda_{A}=8 \lambda, \lambda_{B}=\lambda, N_{0 A}=N_{0 B}=N_{0}$

$ \therefore \quad \frac{N_{B}}{N_{A}} =\frac{e^{-\lambda t}}{e^{-8 \lambda t}} $

$\frac{1}{e}=e^{-\lambda t} e^{8 \lambda t}=e^{7 \lambda t}$

$\Rightarrow-1=7 \lambda t$ or $t=\frac{-1}{7 \lambda}$

$N_{\mathrm{1}}=$ Remaining nuclei after $40 \%$ decay

$=(1-0.4) N_{0}=0.6 N_{0}$

$N_{2}=$ Remaining nuclei after $85 \%$ decay

$=(1-0.85) N_{0}=0.15 N_{0}$

$\therefore$  $\frac{N_{2}}{N_{1}}=\frac{0.15 N_{0}}{0.6 N_{0}}=\frac{1}{4}=\left(\frac{1}{2}\right)^{2}$

Hence, two half life is required between $40 \%$ decay and $85 \%$ decay of a radioactive substance.

$\therefore$ Time taken $=2 \tau_{1 / 2}=2 \times 30 \mathrm{min}=60 \mathrm{min}$

where $R_{o}=1.5 \mathrm{fm}=1.5 \times 10^{-15} \mathrm{m}$

and $A$ is the mass no.

$A^{1 / 3}$ is of order $1$

hence $R$ is of order $10^{-15} \mathrm{m}$

$\therefore {r_2} = 3{\left( {\frac{{206}}{4}} \right)^{1/3}} = 11.6\;Fermi$.

==> $\frac{{{v_1}}}{{{v_2}}} = \frac{8}{1} = \frac{{{m_2}}}{{{m_1}}}$…… $(i) $

Also from $r \propto {A^{1/3}}$ 

==> $\frac{{{r_1}}}{{{r_2}}} = {\left( {\frac{{{A_1}}}{{{A_2}}}} \right)^{1/3}} = {\left( {\frac{1}{8}} \right)^{1/3}} = \frac{1}{2}$.

==> $\frac{3}{5} = {\left( {\frac{{27}}{A}} \right)^{1/3}}$

==> $\frac{{27}}{{125}} = \frac{{27}}{A}$

==> $A = 125$ 

Number of nuclei in atom $X   = A - 52 = 125 - 52 = 73.$

==> $log_e \,R = log_e\, R_0 + \frac{1}{3}{\log _e}A$

This is the equation of a straight line with positive slope.

For ionised Helium atom $Z =2$

Second Bohr orbit $n =2$

$\Rightarrow r _2=\frac{0.529 \times 2^2}{2}=1.06\, \mathring A$

$\therefore q V=\frac{1}{2} m v^{2} \Rightarrow q \times \frac{K(Z e)}{r}=\frac{1}{2} m v^{2}$

$\Rightarrow r=\frac{2(2 e) K(Z e)}{m v^{2}}=\frac{4 K Z e^{2}}{m v^{2}}$

$\Rightarrow r \propto \frac{1}{v^{2}}$ and $r \propto \frac{1}{m}$

$\Rightarrow$ Binding energy per nucleon remains same $\Rightarrow$ No energy is released

$(2)$ For $51 < {\rm{A}} < 100,$ on fusion mass number for compound nucleus is between $100$ and $200$

$\Rightarrow$ Binding energy per nucleon increases

$\Rightarrow$ Energy is released.

$(3)$ For $100 < {\rm{A}} < 200,$ on fission, the mass number of product nuclei will be between $50 $ and $ 100$

$\Rightarrow$ Binding energy per nucleon decreases

$\Rightarrow$ No energy is released

$(4)$ For $200 < {\rm{A}} < 260,$ on fission, the mass number of product nuclei will be between $100 $ and $130$

$\Rightarrow$ Binding energy per nucleon increases

$\Rightarrow$ Energy is released.

$\Rightarrow \frac{3}{5}=\left(\frac{27}{\mathrm{A}}\right)^{1 / 3} \Rightarrow \frac{27}{125}=\frac{27}{\mathrm{A}} \Rightarrow \mathrm{A}=125$

Number of neutrons in atom $\mathrm{X}$

$=A-52=125-52=73$

$\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}=\left(\frac{\mathrm{A}_{1}}{\mathrm{A}_{2}}\right)^{1 / 3}=\left(\frac{64}{125}\right)^{1 / 3}=\frac{4}{5}$

Total Energy initial $=$ Total energy final

Hence $\mathrm{KE}=\frac{\mathrm{K}(\mathrm{ze}) \mathrm{e}}{\mathrm{d}}=\mathrm{PE}$

as $z = 92$      $\boxed{d = 92\frac{{{\text{K}}{{\text{e}}^2}}}{{({\text{KE}})}}}$

Here ${m_e} = 9.1 \times {10^{ - 31}}kg$ and $c =$ velocity of light

$\therefore $Rest energy $ = 9.1 \times {10^{ - 31}} \times {(3 \times {10^8})^2}joule$

$ = \frac{{9.1 \times {{10}^{ - 31}} \times {{(3 \times {{10}^8})}^2}}}{{1.6 \times {{10}^{ - 19}}}}eV = 510\;keV$

It is defined as one twelfth of the mass of an unbound neutral atom of carbon$-12$ in its nuclear and electronic ground state.

Mass of $1$ mole $C$ atoms $= 12gms$

$1 \mathrm{amu}=\frac{12}{12 \times 6.023 \times 10^{23}}=1.66 \times 10^{-27} \mathrm{kg}$

Energy equivalent, $E=m c^{2}=1.66 \times 10^{-27} \mathrm{kg} \times\left(3 \times 10^{8}\right)^{2}=930 \mathrm{MeV}$

$ = \frac{{2.7 \times {{10}^{13}}}}{{3.6 \times {{10}^6}}} = 7.5 \times {10^6}kWh$.

$\Rightarrow 1000 = \frac{{n \times 200 \times {{10}^6} \times 1.6 \times {{10}^{ - 19}}}}{t}$

$\Rightarrow \frac{n}{t}=3.125\times {{10}^{13}}.$

$\therefore E = (\Delta m){c^2} = (0.007 \times {10^{ - 3}}){(3 \times {10^8})^2}$$ = 63 \times {10^{10}}J$.

$ = 1.67 \times {10^{ - 27}}kg = 1amu$

Energy equivalent to $1\;amu = 931\;MeV$

So energy equivalent to $2\, amu = 2 \times 931\;MeV$

$ = 1862 \times {10^6} \times 1.6 \times {10^{ - 19}}$$ = 2.97 \times {10^{ - 10}}J$$ = 3 \times {10^{ - 10}}J$.

$ \Rightarrow E = \frac{{4.5 \times {{10}^{13}}}}{{3.6 \times {{10}^6}}} = 1.25 \times {10^7}kWH.$

==>$\frac{{{E_d}}}{{{E_\alpha }}} = \frac{{{m_\alpha }}}{{{m_d}}}$

==>${E_d} = \frac{{{E_\alpha } \times {m_\alpha }}}{{{m_d}}}$

$ = \frac{{6.7 \times 4}}{{214}} = 0.125\,MeV$

$= \frac{{{m_0}{c^2}}}{{\sqrt {1 - ({v^2}/{c^2})} }} - {m_0}{c^2}$

$ = {m_0}{c^2}\left( {\frac{1}{{\sqrt {1 - ({v^2}/{c^2})} }} - 1} \right) $

$= 0.511\,\left( {\frac{1}{{\sqrt {0.75} }} - 1} \right)$

 $=0.079\, MeV$

$ = 13.6\;{Z^2}\left[ {\frac{1}{{{1^2}}} - \frac{1}{{{2^2}}}} \right]$

$ \Rightarrow 40.8 = 13.6 \times \frac{3}{4} \times {Z^2} \Rightarrow Z = 2.$

Now required energy to remove the electron from ground state 

$ = \frac{{ + 13.6{Z^2}}}{{{{(1)}^2}}} = 13.6{(Z)^2} = 54.4\,eV.$

$K.E. = \left( {\frac{1}{2}m{v^2}} \right) = \frac{3}{2}kT.$

$ \Rightarrow 10.2 \times 1.6 \times {10^{ - 19}} = \frac{3}{2} \times (1.38 \times {10^{ - 23}})T$

==> $T = 7.9 ×10^4K.$