Physics MCQs for NEET — Practice Questions with Answers

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.

In radioactive dating, the age of an object is determined by measuring the remaining amount of a radioactive isotope. If the amount of ${\mathrm{C}}^{14}$ is one-fourth of the initial amount, it means that three half-lives have passed. So, the age of the wood is 3 × 5700 = 11,400 years.

$$\begin{gathered} X(n, \alpha ) {}_3{}^{7}Li\Rightarrow {}_{Z}X_A+{}_{0}n_1\to {}_{3}L{i}^7+{}_{2}H{e}^4 \\ Z=3+2=5 and A=7+4-1=10 \\ \therefore {}_{5}X_{10}={}_{5}B_{10} \end{gathered}$$

Remaining material $N=\frac{N_0}{2^{t/T}}$

$\Rightarrow N=\frac{10}{{\left(2\right)}^{20/15}}=\frac{10}{2.51}=3.96 g$

So decayed material = 10 - 3.96 = 6.04 g

$$\begin{gathered} {}_{Z}P_A\to {}_{Z-2}D_{A-4}, {}_{Z}P_A\to {}_{Z+1}D_A \\ [P s\tan ds for parent \\ D s\tan ds for daughter nucleus] \end{gathered}$$

$\alpha$ decay causes decrease in A by 4 and Z decreased by 2 where as in $\beta$ decay Z increases by 1 and A remains same. In $\gamma$ decay Z and A remian same.

${}_{72}X_{180}\stackrel{\alpha }{\to }{}_{70}X_1^{176}\stackrel{\beta }{\to }{}_{71}X_2^{176}\stackrel{\alpha }{\to }{}_{69}X_3^{172}\stackrel{\gamma }{\to }{}_{69}X_4^{172}$

$$\begin{gathered} X_1=N_0{e}^{-{\lambda }_{1}t}=N_0{e}^{-10\lambda t;} \\ {X}_2=N_0{e}^{-{\lambda }_{2}t}=N_0{e}^{-\lambda t;} \\ \frac{X_1}{X_2}=\frac{N_0{e}^{-10\lambda t}}{N_0{e}^{-\lambda t}}=e^{-9\lambda t}; given \frac{X_1}{X_2}=\frac{1}{e} \\ \therefore \frac{1}{e}=e^{-9\lambda t} or e^{-1}=e^{-9\lambda t}, \therefore 1=9\lambda t \\ or t=\frac{1}{9\lambda } \end{gathered}$$