Physics MCQs for NEET — Practice Questions with Answers

Total time given = 80 min

Number of half-lives of $A, n_A=\frac{80 min}{20 min}=4$

Number of half-lives of B, $n_B=\frac{80 min}{40 min}=2$

Number of nuclei remained undecayed

$N=N_0{\left(\frac{1}{2}\right)}^n$

where $N_0$ is initial number of nuclei and N is number of nuclei

So for two different cases (A) and (B),

$$\begin{gathered} \frac{N_A}{N_B}=\frac{{\left({\displaystyle \frac{1}{2}}\right)}^{n_A}}{{\left({\displaystyle \frac{1}{2}}\right)}^{n_B}} or \frac{N_A}{N_B}=\frac{{\left({\displaystyle \frac{1}{2}}\right)}^4}{{\left({\displaystyle \frac{1}{2}}\right)}^2}=\frac{\left({\displaystyle \frac{1}{16}}\right)}{\left({\displaystyle \frac{1}{4}}\right)} \\ or \frac{N_A}{N_B}=\frac{1}{4} \end{gathered}$$

The radioactive isotope ${}_{27}{}^{60}Co$ decays into the stable isotope ${}_{28}{}^{60}Ni$ by emitting two gamma rays with energies 1.17 MeV and 1.33 MeV in succession.

Order of Radius of atom $\approx {10}^{-10} m$

Order of Radius of nucleus $\approx {10}^{-15} m$

Ratio of volume of atom to volume of nucleus

$$\begin{gathered} =\frac{Volume of atom}{Volume of nucleus} \\ =\frac{{\displaystyle \frac{4}{3}}{\mathrm{\pi r}}_1^3}{{\displaystyle \frac{4}{3}}{\mathrm{\pi r}}_2^3}={\left(\frac{{10}^{-10}}{{10}^{-15}}\right)}^3={10}^{15} \end{gathered}$$

${}_{72}A_{180}\stackrel{\alpha }{\to }{}_{70}A_1^{176}\stackrel{\beta }{\to }{}_{71}A_2^{176}\stackrel{\alpha }{\to }{}_{69}A_3^{172}\stackrel{\gamma }{\to }{}_{69}A_4^{172}$

$$\begin{gathered} Fraction undecayed after two half-lives ={\left(\frac{1}{2}\right)}^2=\frac{1}{4} \\ Fraction decayed after two half-lives=\frac{3}{4} \end{gathered}$$

If R is the radius of the nucleus, the corresponding volume $\frac{4}{3}{\mathrm{\pi R}}^3$ has been found to be proportional to A.

The relationship is expressed in inverse form as R = $R_0{A}^{1/3}$

The value of $R_0$ is $1.2\times {10}^{-15}$ m, i.e, 1.2 fm

$$\begin{gathered} Therefore, \frac{R_{Al}}{R_{Te}}={\left(\frac{A_{Al}}{A_{Te}}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}={\left(\frac{27}{125}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}=\frac{3}{5} \\ {R}_{Te}=\frac{5}{3}\times 3.6=4 fm \end{gathered}$$

$$\begin{gathered} \frac{N}{N_0}={\left(\frac{1}{2}\right)}^{1/T}\Rightarrow \left(\frac{1}{16}\right)={\left(\frac{1}{2}\right)}^{2/T} \\ \Rightarrow {\left(\frac{1}{2}\right)}^4={\left(\frac{1}{2}\right)}^{2/T}\Rightarrow T=0.5 hour=30 minutes \end{gathered}$$

Let $n_1$ and $n_2$ be the number of atoms in ${}_5{}^{10}B and {}_5{}^{11}B$ isotopes.

Atomic weight 

$$\begin{gathered} =\frac{n_1\times (At. wt. of {}_5{}^{10}B)+n_2\times (At. wt. of {}_5{}^{11}B)}{n_1+n_2} \\ or 10.81=\frac{n_1\times 10+n_2\times 11}{n_1+n_2} \\ or 10.81 n_1+10.81 n_2=10 n_1+ 11 n_2 \\ or 0.81 n_1=0.19 n_2 \\ or \frac{n_1}{n_2}=\frac{0.19}{0.18}=\frac{19}{81} \end{gathered}$$

Note:- Atomic weight of an atom having two or more isotopes is the average of the total weight of two of more isotopes.

Number of atoms left after n half-lives is given by

$$\begin{gathered} N=N_0{\left(\frac{1}{2}\right)}^n \\ [N_0=initial count] \\ [N=final count rate of the n half life] \\ or \frac{N}{N_0}={\left(\frac{1}{2}\right)}^n \\ where n=\frac{t}{T_{1/2}} \\ \therefore n=\frac{6400}{1600}=4 \\ [t=6400] \\ [T_{1/2}=1600] \\ \therefore \frac{N}{N_0}={\left(\frac{1}{2}\right)}^4=\frac{1}{16} \end{gathered}$$

The equation for initial and final count rate is

$$\begin{gathered} N=N_0{\left(\frac{1}{2}\right)}^n \\ [N=final count rate] \\ [N_0=initial rate of radio-active atom] \\ where, n=\frac{t}{T_{1/2}} \\ Here, n=\frac{120}{30}=4 \\ [\because t=2h=2\times 60 min=120 min] \\ \therefore \frac{N}{N_0}={\left(\frac{1}{2}\right)}^4=\frac{1}{16} \\ or N_0=16\times N \\ =16\times 5=80 s^{-1} \end{gathered}$$