NEET Practice Questions (MCQs) with Answers & Solutions

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

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

From equation (Ist) there is 1 $\alpha$-decay in which B has atomic no. 2 less than A. In IInd case there is 2-$\beta$-decay in which C has atomic no. 2 greater than B, since A and C  have same atomic no. so they are called isotopes.

$R=R_0{A}^{1/3}, R \alpha A^{1/3}$

$$\begin{gathered} From N=N_0{e}^{-\lambda t} \\ 1800=2000 e^{-\lambda \times 2} \\ \frac{9}{10}=e^{-2\lambda } \\ {e}^{-\lambda }={\left(\frac{9}{10}\right)}^{1/2} \\ Number of nuclei left after 6s \\ N=N_0{e}^{-\lambda t\text{'}} \\ =2000 e^{-\lambda \times 6} \\ Now, puting the value of e^{-\lambda /2} \\ \Rightarrow N=2000\times \frac{729}{1000}=1458 \end{gathered}$$

(a) The decay rate R of a radioactive material is the number of decays per seconds.

From radioactive decaylaw,

      $-\frac{dN}{dt}\propto N$

or    $-\frac{dN}{dt}=\lambda N$

i.e. Rate of reaction is directly proportional to the initial concentration of reactants.

Thus,     $R=-\frac{dN}{dt} or R\propto N$

or         $R=\lambda N or R=\lambda N_0{e}^{-\lambda t}$                     ...(i)

where $R_0=\lambda N_0$ is the activity of the radioactive material at time t=0.

At time $t_1, R_1=R_0 e^{-\lambda t_1}$                      ...(ii)

At time $t_2, R_2=R_0 e^{-\lambda t_2}$           ...(iii)

Dividing Eq. (ii) by Eq. (iii), we have 

       $\frac{R_1}{R_2}=\frac{e^{-\lambda t_1}}{e^{-\lambda t_2}}=e^{-\lambda \left(t_1-t_2\right)}$

or    $R_1=R_2 e^{-\lambda \left(t_1-t_2\right)}$

Binding energy for light nuclei (A < 20) is much smaller than the binding energy for heavier nuclei. This suggests a process that is reverse of fission. This suggests a process that is reverse of fission. When two light nuclei combine to form a heavier nucleus, the process is called nuclear fusion. The union of two light nuclei into heavier nuclei also lead to a transfer of  mass and a consequent liberation of large amount energy.