Physics MCQs for NEET — Practice Questions with Answers

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.

$$\begin{gathered} E=m{c}^2 \\ 1000\times {10}^3\times 3600=m\times {\left(3 \times {10}^8\right)}^2 \end{gathered}$$

Nuclear fusion takes place in stars which results in joining of nuclei accompanied by release of tremendous amount of energy.

According to proton-neutron hypothesis, a nucleus of mass number A and atomic number Z contains Z protons and (A-Z) neutrons. Constituents of atomic nucleus are Nucleons i.e neutron and proton.

Amount of substance remained is 

$$\begin{gathered} M=M_0{\left(\frac{1}{2}\right)}^n \\ [M=subs\tan ce remained] \\ [M_0=initial amount] \end{gathered}$$

Given, $M_0$= 100 g, M = 25 g,

Half life of radioactive substance $T_{1/2}=1600 yr$

$$\begin{gathered} So, 25=100{\left(\frac{1}{2}\right)}^n \\ or \frac{25}{100}={\left(\frac{1}{2}\right)}^n \\ or {\left(\frac{1}{2}\right)}^2={\left(\frac{1}{2}\right)}^n \\ Comparing the power, we have \\ n=2 \\ or \frac{t}{T_{1/2}}=2 \\ or t=2T_{1/2}=2\times 1600 \\ =3200 yr \end{gathered}$$

The reaction can be written as

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

The energy released in the reaction is the difference of binding energies of daughter and parent nuclei. Hence, energy released 

= binding energy of ${}_{2}H{e}^4-2\times binding energy of {}_{1}H_2$

= $28-2\times 2.2=23.6 MeV$