NEET Practice Questions (MCQs) with Answers & Solutions

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

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)] $c^2$

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

Let the unknown product nucleus be ${}_{Z}X_A$

The reaction can be writted as

$$\begin{gathered} {}_{8}O_{16} + {}_{1}H_2 \to {}_{z}X_A + {}_{2}H{e}^4 \\ \left(oxygen\right) \left(deuterium\right) (unknown \alpha -particle \\ nucleus) \end{gathered}$$

Conservartion of mass number between product and reactant of above reaction gives.

16 + 2 = A + 4 $\Rightarrow$A = 14

Conservation of atomic number between reactant and product of above reaction gives

8 + 1 = Z + 2 $\Rightarrow$Z = 7

Thus, the unknown product nucleus is nitrogen $\left({}_{7}N_{14}\right)$

Note:- Fusion reaction can take place at very high temperature ($\approx {10}^8$ K) and very high pressure which can be provided at sun or by fission of atom bomb.

Since in the given reaction ${}_{-1}e_0$ and antineutrino $\left(\overline{\nu }\right)$ are released, so it can be considered $\beta$-decay.

$M\left({}_{7}N_{15}\right)+1 m_p \to M\left({}_{8}O_{16}\right)$

binding energy of last proton

$$\begin{gathered} =\left[M\left(N^{15}\right)+m_p-M\left({}_{1}O_{16}\right)\right]\times 931.5 MeV \\ =\left[15.00011+1.00783-15.99492\right] \times 931.5 MeV \\ =0.01302 \times 931.5 MeV=12.13 MeV \end{gathered}$$

Radioactivity is a nuclear property and cannot be controlled by external factors.

We have the energy released by fission of one uranium atom is 200 MeV.

So, $E=200\times {10}^6\times 1.6\times {10}^{-19}$

           = 3.2$\times {10}^{-11} J$

The power required = 3.2 W

Thus the number of fission required is equal to $\frac{3.2}{3.2\times {10}^{-11}}={10}^{11}$ fissions.