⇒Structure of Atom and Nuclei

Correct Option is (C)

Given:

Total time, $\mathrm{T}=10$ years

Half life, ${\mathrm{T}}_{1/2}=5$ years

$\therefore$ No. of half lives, $\mathrm{n}=\frac{10}{5}=2$

From $\frac{\mathrm{N}}{{\mathrm{N}}_0}={\left(\frac{1}{2}\right)}^{\mathrm{n}}$

$\therefore \frac{\mathrm{N}}{{\mathrm{N}}_0}={\left(\frac{1}{2}\right)}^2=\frac{1}{4}$

$\therefore$ After 10 years, ${\frac{1}{4}}^{\text{th }}$ of the original substance will remain.

$\Rightarrow (1-\frac{1}{4})=\frac{3}{4}\times 100=75\mathrm{\%}$ of the fraction would get decayed.

Correct Option is (B)

${}_{\mathrm{Z}}{\mathrm{X}}^{\mathrm{A}}\stackrel{\alpha }{⟶}{}_{\mathrm{Z}-2}{\mathrm{Y}}^{\mathrm{A}-4}\stackrel{2{\beta }^{-}}{⟶}{}_{\mathrm{Z}}{\mathrm{X}}^{\mathrm{A}-4}$

Correct Option is (C)

$\therefore$ Activity is given as:

$\begin{array}{rl}\therefore \mathrm{A}& =\lambda \mathrm{N}\\ \lambda & =\frac{\ln 2}{\mathrm{T}}\end{array}$

$\therefore$ Ratio of two different radioactive elements will be:

$\begin{array}{rl}& \frac{{\mathrm{A}}_1}{{\mathrm{A}}_2}=\frac{{\lambda }_1{\mathrm{N}}_1}{{\lambda }_2{\mathrm{N}}_2}\\ & \frac{{\mathrm{A}}_1}{{\mathrm{A}}_2}=\frac{\frac{\ln 2}{{\mathrm{T}}_1}{\mathrm{N}}_1}{\frac{\ln 2}{{\mathrm{T}}_2}{\mathrm{N}}_2}\\ & \frac{{\mathrm{A}}_1}{{\mathrm{A}}_2}=\frac{{\mathrm{N}}_1{\mathrm{T}}_2}{{\mathrm{N}}_2{\mathrm{T}}_1}\end{array}$

Correct Option is (B)

According to radioactive disintegration law,

$\begin{array}{rl}& \mathrm{N}={\mathrm{N}}_0{\mathrm{e}}^{-\lambda \mathrm{t}}\\ & \frac{\mathrm{N}}{{\mathrm{N}}_0}={\mathrm{e}}^{-\lambda \mathrm{t}}\\ & \frac{\mathrm{N}}{{\mathrm{N}}_0}={\mathrm{e}}^{-\lambda \times \frac{1}{2\lambda }}\dots (\because \mathrm{t}=\frac{1}{2\lambda })\\ & \frac{\mathrm{N}}{{\mathrm{N}}_0}={\mathrm{e}}^{-\frac{1}{2}}\\ & \frac{{\mathrm{N}}_0}{\mathrm{N}}={\mathrm{e}}^{\frac{1}{2}}\\ & \frac{{\mathrm{N}}_0}{\mathrm{N}}=\sqrt{\mathrm{e}}\end{array}$

Correct Option is (A)

Half life, $T=15$ days, time $t=60$ days $=4T$

The number of nuclei remaining is given by

$\begin{array}{rl}& \mathrm{N}={\mathrm{N}}_0{\left(\frac{1}{2}\right)}^{\mathrm{n}}={\mathrm{N}}_0{\left(\frac{1}{2}\right)}^4\mathrm{n}=\text{ no. of half-lives }\\ & =\frac{1}{16}{\mathrm{N}}_0=\frac{1}{16}\times 8\times {10}^{16}\\ & =0.5\times {10}^{16}\\ & \therefore \text{ No. of nuclei decayed }={\mathrm{N}}_0-\mathrm{N}\\ & =8\times {10}^{16}-0.5\times {10}^{16}=7.5\times {10}^{16}\end{array}$