NEET Practice Questions (MCQs) with Answers & Solutions

In sun, huge amount of energy is produced due to fusion of 4 protons (hydrogen nucleus) into a helium nucleus. According to the reaction

${}_{1}H_1+{}_{1}H_1+{}_{1}H_1+{}_{1}H_1\to {}_{2}H{e}^4+2{}_{+1}\beta _0+\gamma \left(energy\right)+2\nu$

Number of half-lives

$$\begin{gathered} n=\frac{t}{T_{1/2}}=\frac{30 days}{10 days}=3 \\ [T_{1/2}=half life period] \end{gathered}$$

So, number of undecayed radioactive nuclei is given

$$\begin{gathered} \frac{N}{N_0}={\left(\frac{1}{2}\right)}^n \\ [N=Final number] \\ [N_0=Initial number] \\ or N=N_0{\left(\frac{1}{2}\right)}^n=4\times {10}^{10}{\left(\frac{1}{2}\right)}^3 \\ =4\times {10}^{10}\times \frac{1}{8}=0.5\times {10}^{10} \\ Number of nuclei decayed after 30 days \\ =N_0-N \\ =4\times {10}^{10}-0.5\times {10}^{10} \\ =3.5\times {10}^{10} \end{gathered}$$

$$\begin{gathered} As, N=N_0{\left(\frac{1}{2}\right)}^{t/T} \\ \therefore N_A=10{\left(\frac{1}{2}\right)}^{t/1} \\ {N}_B=1{\left(\frac{1}{2}\right)}^{t/2} \\ Given, N_A=N_B \\ So, 10{\left(\frac{1}{2}\right)}^t={\left(\frac{1}{2}\right)}^{t/2} \\ So, 10={\left(\frac{1}{2}\right)}^{-t/2} \\ or 10=2^{t/2} \\ \therefore {\log }_{10} 10=\frac{t}{2} {\log }_{10} 2 \\ 1=\frac{t}{2}\times 03010 \\ \therefore t=6.62 yr \end{gathered}$$

(b) Beta decay involves the emission of either electrons or positrons. The electrons or positrons emitted in a $\beta$-decay do not exist inside the nucleus. They are only created at the time of emission, just as photons are created when an atom makes a transition from higher to a lower energy state.

In negative $\beta$-decay a neutron in the nucleus is transformed into a proton, an electron and an antineutrino. Hence, in radioactive decay process, the negatively charged emitted $\beta$-particles are the electrons produced as a result of the decay of neutrons present inside the necleus.

Since mean life period is $t_{mean}=\frac{1}{\lambda }$

2 means lives = $\frac{2}{\lambda }$

Also by radioactive decay equation, $N=N_0{e}^{-\lambda }$

$$\begin{gathered} \therefore N=N_0{e}^{-\lambda \frac{2}{\lambda }}=N_0{e}^{-2}=\frac{N_0}{e^2} \\ =\frac{N_0}{{(2.718)}^2}=\frac{N_0}{7.39}=0.135 N_0 \\ Also M=\frac{M_0}{e^2}=\frac{M_0}{7.39}=0.135 M_0 \\ =0.135\times 10=1.35 gm \end{gathered}$$

$$\begin{gathered} N=N_0{\left(\frac{1}{2}\right)}^2\Rightarrow \frac{N}{N_0}=\frac{1}{4} \\ Probability=1-\frac{N}{N_0}=1-\frac{1}{4}=\frac{3}{4} \end{gathered}$$

${}_{92}U_{235}+{}_{0}n_1\to {}_{38}S{r}^{90}+{}_{54}X{e}^{143}+3{}_{0}n_1$

If total atomic number on LHS = 92 + 0 = 92

Total atomic number on RHS = 38 + 54 + 0 = 92

Total mass number on LHS = 235 + 1 = 236

Total mass number on RHS = 90 + 143 + 3 $\times$ 1 = 236

Note :- For a nuclear reaction to be completed,the mass number and charge number on both sides should be same.

(a) Number of half-lives

        $n=\frac{t}{T_{1/2}}=\frac{18}{3}=6 \left[T_{1/2} = half life period\right]$

Amount remained after n half-lives

       $N=N_0{\left(\frac{1}{2}\right)}^n \left[\begin{array}{c}{N}_0=initial count\\ N=final count\end{array}\right]$

Given,   $N_0=300g$

$\therefore N=300{\left(\frac{1}{2}\right)}^6=300\times \frac{1}{64}=4.68 g$

$$\begin{gathered} Decay cons\tan t, \lambda =\frac{0.693}{T_{1/2}} \\ =\frac{0.693}{T_{1/2}}=\frac{0.693}{20}=0.03465 \\ Now, time of decay, \\ t=\frac{2.303}{\lambda } \log \frac{N_0}{N} \\ \Rightarrow t_1=\frac{2.303}{0.03465} \log \frac{100}{67}=11.6 min \\ and t_2=\frac{2.303}{0.03465} \log \frac{100}{33}=32 min \\ So, time difference between points of time \\ =t_1-t_2=32-11.6 \\ =20.4 min\approx 20 min \end{gathered}$$