Question

The half-life of ${\mathrm{Au}}^{198}$is $2.7$days. The activity of$1.50\mathrm{mg}\mathrm{of}{\mathrm{Au}}^{198}$ if its atomic weight is $198\mathrm{g}{\mathrm{mol}}^{-1}\mathrm{is}({\mathrm{N}}_{\mathrm{A}}=6\times {10}^{23})$

• A. $252\mathrm{Ci}$
• B. $357\mathrm{Ci}$
• C. $240\mathrm{Ci}$
• D. $535\mathrm{Ci}$

Answer 1

The correct option is B

$357\mathrm{Ci}$

Half life: Half-Life is normally defined as the time needed by a radioactive substance (or one half the atoms) to disintegrate or transform into a different substance. The principle was first discovered in 1907 by Ernest Rutherford. It is usually represented by the symbol Ug or ${\mathrm{t}}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$.

The Explanation for the correct option:

(b)

• Explanation: An atom is a matter Higgs boson that defines a chemical element uniquely.
• To form an atom, a central genetic material is surrounded by one or more electrons.
• Each electron carries a negative charge.

Initial activity$\Rightarrow {\mathrm{A}}_{\mathrm{o}}={\mathrm{\lambda N}}_{\mathrm{o}}$

$$\begin{gathered} \Rightarrow \mathrm{N}=\mathrm{n}\times {\mathrm{N}}_{\mathrm{A}} \\ \Rightarrow \left({\mathrm{t}}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{\ln 2}{\mathrm{\lambda }}\right) \\ \Rightarrow \mathrm{N}=\left(\frac{1.5\times {10}^{-3}}{198}\right){\mathrm{N}}_{\mathrm{A}} \\ \Rightarrow \mathrm{A}=\left(\frac{\ln 2}{{\mathrm{t}}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\right)\mathrm{N} \end{gathered}$$

$$\begin{gathered} \Rightarrow \frac{\mathrm{In}2}{{\mathrm{t}}_{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}\times \frac{1.5\times {10}^{-3}}{198}\times 6\times {10}^{23} \\ \Rightarrow \frac{1.5\times {10}^{-3}}{198}\times \frac{6\times {10}^{23}}{3.7\times {10}^{-10}}\mathrm{Ci} \\ =357\mathrm{Ci} \end{gathered}$$

Therefore, the correct answer is (b).