The half-life of ${}_{38}{}^{90}\mathrm{Sr}$ is 28 years. What is the disintegration rate of 15 mg of this isotope?

Given,
Half- life of ${}_{38}{}^{90}\mathrm{Sr}$ = 28 years. 

Using the formula,
                           $\boxed{\mathrm{\lambda } =\frac{0.693}{\mathrm{T}}}$

$\Rightarrow$                       $\mathrm{\lambda } = \frac{0.693}{28\times 365\times 24\times 60\times 60}$
                                  
                              $= 7.85 \times {10}^{-10} {\mathrm{s}}^{-1}$ 
90 g of Sr contains 6.023 x 10²³ atoms.

$\therefore$ 15 mg of Sr contains,

${\mathrm{N}}_0 = \frac{6.023 \times {10}^{23} \times 15 \times {10}^{-3}}{90}\mathrm{atoms}$ 

${\mathrm{N}}_0 = 1.0038 \times {10}^{20} \mathrm{atoms}$

Disintegration rate, $\boxed{\frac{\mathrm{dN}}{\mathrm{dt}} = -\mathrm{\lambda } {\mathrm{N}}_0}$

                                       $$\begin{gathered} = -7.85 \times {10}^{-10} \times 1.0038 \times {10}^{20} \\ = 7.88 \times {10}^{10} \mathrm{dps} \mathrm{or} \mathrm{Bq} \\ = \frac{7.88 \times {10}^{10}}{3.7 \times {10}^{10}}\mathrm{Ci} \end{gathered}$$

                                    $= 2.13 \mathrm{Ci}.$

Given, the sample has a half life of T years. 

(a) The fraction of the original sample left is ,
                 $\frac{\mathrm{N}}{{\mathrm{N}}_{\mathrm{o}}}$= $\frac{3.125}{100}$
                      $= \frac{1}{32}= {\left(\frac{1}{2}\right)}^5$

           
Hence,  there are 5 half lives of T years spent. Thus, the time taken is 5T years.

(b) The fraction of the original sample left = $\frac{1}{100}={\left(\frac{1}{2}\right)}^{\mathrm{n}}$ 

 i.e.,     $2^{\mathrm{n}}$ = 100  

$\Rightarrow$  n log 2 = log 100 

Hence, n =$\frac{\log 100}{\log 2}=\frac{2}{0.301}$  = 6.64 

From, t= nT we have, t = 6.64 T.

Hence, there are 6.64 half lives of T years spent. Thus, the time taken is 6.64T years.

(i) ${}_{88}{}^{226}\mathrm{Ra} \to {}_{86}{}^{222}\mathrm{Rn} + {}_2{}^4\mathrm{He}$

(ii) ${}_{94}{}^{242}\mathrm{Pu} \to {}_{92}{}^{238}\mathrm{U} + {}_2{}^4\mathrm{He}$

(iii) ${}_{15}{}^{32}\mathrm{P} \to {}_{16}{}^{32}\mathrm{S} + {\mathrm{e}}^{-1} + \overline{\mathrm{v}}$

(iv) ${}_{83}{}^{210}\mathrm{Bi} \to {}_{84}{}^{210}\mathrm{Po} + {\mathrm{e}}^{-1}+\overline{\mathrm{v}}$

(v)  ${}_6{}^{11}\mathrm{C} \to {}_5{}^{11}\mathrm{B} + {\mathrm{e}}^{+}+\mathrm{v}$

(vi) ${}_{43}{}^{97}\mathrm{Tc} \to {}_{42}{}^{97}\mathrm{Mo} + {\mathrm{e}}^{+} +\mathrm{v}$

(vii) ${}_{54}{}^{120}\mathrm{Xc} + {\mathrm{e}}^{-} \to {}_{53}{}^{120}\mathrm{I} + \mathrm{v}$