Physics MCQs for NEET — Practice Questions with Answers

$\lambda =\left({\lambda }_1+{\lambda }_2\right)=\frac{1}{1620}+\frac{1}{520}=2.54\times {10}^{-3} Y{r}^{-1}$

$\therefore t_{1/2}=\frac{In\left(2\right)}{\lambda }=272.8 yrs.$

$\frac{1}{4}$th of the material remains intact after 2 half-lives.

$\because \frac{1}{64}={\left(\frac{1}{2}\right)}^6$

$\therefore$ 6 half life period = 60 sec. or Half life period is 10 sec.

${}_{84}P{o}^{210}\to {}_{82}X_{206}+{}_{2}H{e}^4$

Using conservation of linear momentum

206$\nu$'+4$\nu$= 0$\Rightarrow$$\nu \text{'}=-\frac{4\nu }{\left(206\right)}\Rightarrow \left|\nu \text{'}\right|=\frac{4\nu }{206}$

88. ${}_{7}X_{15}+\alpha -particle\to {}_{9}Y_{19}-{}_{1}H_1\to {}_{8}Z_{18}$

(c)      $r\propto A^{1/3}\Rightarrow \frac{r_1}{r_2}={\left(\frac{A_1}{A_2}\right)}^{1/3}$

        $\Rightarrow \frac{3.6}{r_2}={\left(\frac{27}{125}\right)}^{1/3}=\frac{3}{5}\Rightarrow r_2=6 fermi$

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, $N=N_0{e}^{-\lambda }$

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