NEET Practice Questions (MCQs) with Answers & Solutions

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

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.

(d) Mass defect $∆\mathrm{m}=\frac{2.23}{931}$ = 0.0024.

(b) Positron is the antiparticle of electron.

(b) Size of the nucleus is of the order of ${10}^{-15 }\mathrm{m}$

(c) Nuclear force works in the range of ${10}^{-15} \mathrm{m}$

(a) B.E. = $∆{\mathrm{mc}}^2=\left[2\left(1.0087+1.0073\right)-4.0015\right]\times 931.5=28.4 \mathrm{MeV}$