NEET Practice Questions (MCQs) with Answers & Solutions

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

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

= $28-2\times 2.2=23.6 MeV$

The mass of radioactive substance remained is,

$M=M_0{\left(\frac{1}{2}\right)}^n$

Here, final mass, M = 1 g, initial mass, $M_0$= 256 g, half life period, $T_{1/2}$= 12.5 h

$$\begin{gathered} so, 1=256{\left(\frac{1}{2}\right)}^n \\ or \frac{1}{256}={\left(\frac{1}{2}\right)}^n \\ or {\left(\frac{1}{2}\right)}^8={\left(\frac{1}{2}\right)}^n \\ comparing the powers on both the sides, we get \\ n=8=\frac{t}{T_{1/2}} \\ \therefore t=8T_{1/2} \\ =8\times 12.5 \\ =100 h \end{gathered}$$

$$\begin{gathered} \frac{1}{64}={\left(\frac{1}{2}\right)}^n \\ n=6 \end{gathered}$$

6 half-lives are equal to 60 second.$\therefore 1 half-life=10 s$

A nuclear reaction represents the transformation of one stable nucleus into another nucleus by bombarding the former with suitable high energy particles.

${}_{6}C_{12}+{}_{0}n_1\to {}_{6}C_{13}\to {}_{7}N_{13}+{}_{-1}\beta _0+Q \left(energy\right)$

Resulting nucleus is of nitrogen having mass no. 13 and atomic no. 7

In a nuclear fusion, when two light nuclei of different masses are combined to form a stable nucleus, then some mass is lost and appears in the form of energy, called the mass defect. So, the mass of resultant nucleus is always less than the sum of masses of initial nuclei i.e.,

$m_3<\left(m_1+m_2\right)$

The nuclei which have same number of neutrons but different atomic number and mass number are known as isotones. In choice (1) nuclei of ${}_{34}S{e}^{74}$ and ${}_{31}G{a}^{71}$ are isotones as

A - Z = 74 - 34 = 71 - 31 = 40