Physics MCQs for NEET — Practice Questions with Answers

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 = $0.02554\times 931.5 MeV=23.79 MeV$

${}_{1}H_2+{}_{1}H_2\to {}_{2}H{e}^4+energy$

Binding energy of a $\left({}_{1}H_2\right)$ deuterium nuclei

$=2\times 1.1=2.2 MeV$

Total binding energy of two deuterium nuclei

$=2.2\times 2=4.4 MeV$

Binding energy of a $\left({}_{2}H{e}^4\right)$ nuclei = $4\times 7=28 MeV$

So, energy released in fusion = 28 - 4.4 = 23.6 MeV

Binding energy of fissioned nucleus

$=236\times 7.6 MeV$

Binding energy of products

$$\begin{gathered} =117\times 8.5+117\times 8.5 \\ =2\times 117\times 8.5 \end{gathered}$$

Hence, net binding energy = binding energy of products - binding energy of fissioned nucleus

$$\begin{gathered} =234\times 8.5-236\times 7.6 \\ =1989-1793.6 \end{gathered}$$

=195.4 MeV

$\approx 200 Mev$

Thus, in per fission of uranium nearly 200 MeV energy is released.

Q = (1.002 + 1.004 - 1.001 - 1.003) (931.5) MeV

    = 1.863 MeV

Let Z be ;the  charge number and A be the mass number of particle X, then conservation of charge number gives 

6 = 5 + 1 + Z $\Rightarrow$Z = 0

Conservation of mass number gives,

11 = 11 + 0 + A

$\Rightarrow$A = 0

X is a particle of zero charge and zero mass. This particle may be  neutrino or antineutrino. As we know that for positive $\beta$-particle, neutrino is emitted and with negative $\beta$-particle, antineutrino is emitted.

Thus, in this case neutrino will be emitted.

Binding energy of a nucleus containing N neutrons and Z protons is

BE = [N$m_n$ + Z$m_p$ - m(N, Z)] $c^2$

$$\begin{gathered} \Rightarrow \frac{BE}{c^2}=N{m}_n+Z{m}_p-m(N, Z) \\ \Rightarrow m(N, Z)=N{m}_2 + Z{m}_p - BE/c^2 \end{gathered}$$