Physics MCQs for NEET — Practice Questions with Answers

Fusion reaction takes place at high temperatures because kinetic energy is high enough to overcome the Coulomb repulsion between nuclei.

Initially $P\to 4N_0$

          $Q\to N_0$

Half life $T_P\to 1 min$

           $T_Q\to 2min$

Let after time t number of nuclei of P and Q are equal i.e,$\frac{4N_0}{2^{1/1}}=\frac{N_0}{2^{1/2}}$

$4=2^{1/2}$

$2^2=2^{1/2}$

$\frac{t}{2}=2$

t=4min

Disactive nucleus or Nuclei of R

$=\left(4N_0-\frac{4N_0}{2^4}\right)+\left(N_0-\frac{N_0}{2^2}\right)$

$=4N_0-\frac{N_0}{4}+N_0-\frac{N_0}{4}$

$=5N_0-\frac{N_0}{2}$$=\frac{9}{2}{N}_0$

If m=1u, c=$3\times {10}^{8}m{s}^{-1},then$

E=931 MeV ie, 1u=931 MeV

Binding energy=0.042$\times$931=39.10 MeV

$\therefore$Binding energy per nucleon

     $=\frac{39.10}{7}=5.58=5.6 MeV$

Fraction remains after n half lives 

     $\frac{N}{N_0}={\left(\frac{1}{2}\right)}^n={\left(\frac{1}{2}\right)}^{t/T}$

Given       $N=\frac{N_0}{e}\Rightarrow \frac{N_0}{e{N}_0}={\left(\frac{1}{2}\right)}^{5/T}$

or           $\frac{1}{e}={\left(\frac{1}{2}\right)}^{5/T}$

Taking log on both sides, we get 

$\log 1-\log e =\frac{5}{T}\log \frac{1}{2}$

$-1=\frac{5}{T}\left(-\log 2\right)$

$\Rightarrow T=5{\log }_e 2$

Now, let $t^{\text{'}}$ be the time after which activity reduces to half 

           $\left(\frac{1}{2}\right)={\left(\frac{1}{2}\right)}^{t^{\text{'}}/5{\log }_{e}2}$

$\Rightarrow t^{\text{'}}=5{\log }_{e}2$

$$\begin{gathered} {\mathrm{A}}_1=\mathrm{\lambda }{N}_1 \\ {\mathrm{A}}_2=\mathrm{\lambda }{N}_2 \\ N_1-N_2=\left[\frac{{\mathrm{A}}_1-{\mathrm{A}}_2}{\mathrm{\lambda }}\right] \end{gathered}$$

$$\begin{gathered} \mathrm{Mass} \mathrm{of}{ }_1{\mathrm{H}}^2=2.01478 \mathrm{amu} \\ \mathrm{Mass} \mathrm{of}{ }_2{\mathrm{He}}^4=4.00388 \mathrm{amu} \\ \mathrm{Mass} \mathrm{of} \mathrm{two} \mathrm{deuterium}=2\times 2.01478 \\ =4.02956 \mathrm{amu} \\ \mathrm{Energy} \mathrm{equivalent} \mathrm{to}{ }_2{\mathrm{H}}^2 \\ =4.02956\times 1.2 \\ =4.4 \mathrm{MeV} \\ \mathrm{Energy} \mathrm{equivalent} \mathrm{to}{ }_2{\mathrm{He}}^4 \\ =4.00388\times 7 \\ =28 \mathrm{MeV} \\ \mathrm{Energy} \mathrm{released}=(28-4.4) \\ =23.6 \mathrm{MeV} \end{gathered}$$

Let the radioactive substance be ${}_{\mathrm{Z}}{}^{\mathrm{A}}\mathrm{X}$

Radioactive transition is given by 

                           ${}_{\mathrm{Z}}{}^{\mathrm{A}}\mathrm{X}\stackrel{ -\mathrm{a} }{\to }{}_{\mathrm{Z}-2}{}^{\mathrm{A}-4}\mathrm{X}\stackrel{ -2\mathrm{\beta } }{\to }{}_{\mathrm{Z}}{}^{\mathrm{A}-4}\mathrm{X}$

The atoms of element having same atomic number but different mass numbers are called isotopes.

So ${}_{\mathrm{Z}}{}^{\mathrm{A}}\mathrm{X}$ and ${}_{\mathrm{Z}}{}^{\mathrm{A}-4}\mathrm{X}$ are isotopes.

If N is the number of radioactive nuclei present at some instant, then

        $N=N_0{e}^{-\lambda t}$

The constant $N_0$ represents the number of radioactive nuclei at t=0

Now,            $\frac{N_1}{N_2}=\frac{e^{-{\lambda }_{1}t}}{e^{-{\lambda }_{2}t}}$

or              $\frac{N_1}{N_2}=\frac{e^{-5\lambda t}}{e^{-\lambda t}}=e^{-4\lambda t}$

but          $\frac{N_1}{N_2}=\frac{1}{e} \left(as provided\right)$

Therefore,    $\frac{1}{e}=\frac{1}{e^{4\lambda t}}$

or           $4\lambda t=1$

or           $t=\frac{1}{4\lambda }$

Density of nuclear matter is indepent of mass number, so the required ratio is 1:1.

Alternative:

                  $A_1:A_2=1:3$

Their radii will be in the ratio 

             $$\begin{gathered} R_0{A}_1^{1/3}:R_0{A}_2^{1/3}=1:3^{1/3} \\ Density= \frac{A}{{\displaystyle \frac{4}{3}}\pi R^3} \\ \therefore P A_1 : P A_2=\frac{1}{{\displaystyle \frac{4}{3}}\pi R_0^3.1^3}=\frac{3}{{\displaystyle \frac{4}{3}}\pi R_0^3{\left(3^{1/3}\right)}^3} \end{gathered}$$

Their nuclear densities will be the same .

(d) B.E. = $∆$m amu = $∆$m$\times$931 MeV.