Question

The half life of ${}_{92}^{238}U$ undergoing $\alpha$ - decay is $4.5\times {10}^9$ years. Its activity of $1g$ sample is:-

• A. $1.23\times {10}^{4}Bq$
• B. $2.4\times {10}^{5}Bq$
• C. $1.82\times {10}^{6}Bq$
• D. $4.02\times {10}^{8}Bq$

Answer 1

The correct option is A $1.23\times {10}^{4}Bq$

$\begin{array}{cc}\text{Given-}{\ast }_{92}{U}^{238}\text{going}\alpha \text{-decay with}& \\ \text{half life}{t}_{1/2}& =4.5\times {10}^9\text{years.}\end{array}$

$\begin{array}{c}\text{We know that Activity (A),}\\ A=N_0\cdot \lambda \cdot e^{-\lambda t}Bq\end{array}$

$\begin{array}{c}\text{Instant at}t=0\\ \text{we have,}A=N_0\cdot \lambda ---\left(1\right)\\ \ast \text{Calculating (No) -}\\ \text{mol of}\lg {\text{of}}_{92}{U}^{238}=\frac{1}{238}mol\\ \Rightarrow \text{Nucleides}=\frac{N_A}{238}mol-\left(2\right),N_A\to \text{avogadro num.}\end{array}$

$\begin{array}{c}\text{* we know that -}\\ \lambda =\frac{\ln 2}{t_{112}}-\left(3\right)\\ \text{From eqn}\left(1\right),\left(2\right),\text{(3) we get-}\\ A=\frac{6.\times {10}^{23}}{238}\times \frac{42}{4.5\times {10}^9\times 3600\times 24\times 365}Bq\\ A=1.23\times {10}^{4}Bq\\ \text{Hence, opt (A)}\text{is correct.}\end{array}$