NEET Practice Questions (MCQs) with Answers & Solutions

$$\begin{gathered} R=R_0{\left(A\right)}^{1/3} \\ {R}_{Al}=R_0{\left(27\right)}^{1/3}=3R_0 \\ {R}_{Te}=R_0{\left(125\right)}^{1/3}=5R_0=\frac{5}{3}{R}_{Al} \end{gathered}$$

When a nucleus emits an alpha particle, its mass number decreases by 4 and charge/atomic no. decreases by 2. In $\beta$-particle emission, mass remains same but atomic number is increased by one. In $\gamma$-decay, daughter nucleus has the same charge number and same mass number as those of parent nucleus. Hence, sequence is 

${}_{Z}X_A\stackrel{\beta }{\to }{}_{Z+1}Y_A\stackrel{\alpha }{\to }{}_{Z-1}K_{A-4}\stackrel{\gamma }{\to }{}_{Z-1}K_{A-4}$

Mass of proton = mass of antiproton

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

Energy equivalent to 1 amu = 931 MeV

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

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

The relation between nuclei radius (R) and mass number (A) is given  by

$$\begin{gathered} R \alpha A^{1/3}........\left(i\right) \\ or A \alpha R^3 \\ or \frac{A_1}{A_2}={\left(\frac{R_1}{R_2}\right)}^3 \\ Given, R_1=R,R_2=\frac{R}{2}, A=56 \\ \therefore \frac{56}{A_2}={\left(\frac{R}{R/2}\right)}^3=2^3=8 \\ or A_2=\frac{56}{8}=7 \\ Thus, required stable nucleus will be L{i}^7 \end{gathered}$$

Density of nuclear matter is the ratio of mass of nucleus and its volume. If m is average mass of a nucleon and R is the nuclear radius, then mass of nucleus = mA, where A is the mass number of the element.

$Volume of nucleus=\frac{4}{3}\pi R^3$

_{$And R=R_0{\left(A\right)}^{1/3}$}

_{$\Rightarrow V=\frac{4}{3}\pi {\left(R_0\right)}^{3}A$}

As density of nuclear matter = $\frac{mass of nucleus}{volume of nucleus}$

$$\begin{gathered} \rho =\frac{mA}{{\displaystyle \frac{4}{3}}{\mathrm{\pi R}}_0^3\mathrm{A}} \\ \therefore \rho =\frac{3m}{4\pi R_0^3} \end{gathered}$$

As m and $R_0$ are constants, therefore density $\rho$ of nuclear matter is constant.

Fission rate = $\frac{total nuclear power}{energy produced/fission}$

Here, total nuclear power = 5  W

Energy released per fission = 200 MeV

$$\begin{gathered} \therefore Fission rate=\frac{5}{200 MeV} \\ =\frac{5}{200\times 1.6\times {10}^{-13}} \\ [\because 1 Mev=1.6\times {10}^{-13} J] \\ =1.56\times {10}^{11} s^{-1} \end{gathered}$$

The $\alpha$-particle can be represented as ${}_{2}H{e}^4$ and $\beta$-particle as ${}_{-1}\beta _0$. So, after emission of one $\alpha$-particle the mass number of resultant nucleus decreases by 4 unit and atomic number by 2 unit. Similarly, after emission of one $\beta$-particle the atomic number increases by 1 unit keeping its mass number same. So, according to reaction (assuming ${}_{Z}X_A$ the initial nucleus). ${}_{Z}X_A\to {}_{Z-2}Y_{A-4}+{}_{2}H{e}^4 (\alpha -Particle)$ and ${}_{Z-2}Y_{A-4}\to {}_{Z}X_{A-4}+2\left({}_{-1}\beta _0\right) (2 \beta -particles)$

so, by one $\alpha$ and two $\beta$-emissions the atomic number remains unchanged i.e. formation of isotopes takes place.

A moderator in a nuclear reactor is used to slow down the fast-moving neutrons.  Heavy water, graphite or beryllium oxide are used as moderators. Heavy water is the best moderator.

Note:- In an ordinary uranium reactor, plutonium $\left(P{u}^{239}\right)$ is produced which is a better fissionable material than uranium $\left(U^{235}\right)$. It is a heavy isotope of uranium.

The function of a moderator is to slow down the fast moving secondary neutrons produced during the fission as fission reaction can only be initiated by slow moving neutrons.

The material of moderator should be light and it should not absorb neutrons. Usually, heavy water, graphite, deuterium, paraffin etc. Can act as moderators. These moderators are rich in protons.

Mass defect $∆m=2\times 2.01471-4.00388=0.02554$

Energy liberated = $0.02554\times 931.5 MeV=23.79 MeV$