An experiment requires minimum beta activity product at the rate of beta particles per minute. The h — Nuclear Chemistry and Radioactivity Chemistry Question
Question
An experiment requires minimum beta activity product at the rate of $5.0 \times 10^{15}$ beta particles per minute. The half-life period of ${}_{42}^{99}\text{Mo}$, which is a beta emitter is 69.3 h. The minimum moles (in the order $10^{-7}$) of ${}_{42}^{99}\text{Mo}$ required to carry out the experiment in 6.93 h is $[(1.07)^{10} = 2, \log 2 = 0.3, N_A = 6 \times 10^{23}]$
💡 Solution & Explanation
The activity at the end of the 6.93 h experiment must be $A_t = 5.0 \times 10^{15} \text{ dpm}$. The time $t = 6.93 \text{ h}$ is exactly 0.1 half-lives ($6.93 / 69.3$). Thus, $A_t = A_0 (1/2)^{0.1}$. Given $(1.07)^{10} = 2 \implies 2^{0.1} = 1.07$. So, $A_0 = 1.07 \times A_t = 5.35 \times 10^{15} \text{ dpm}$. The decay constant $\lambda = \frac{0.693}{69.3 \times 60} = \frac{1}{6000} \text{ min}^{-1}$. Initial atoms $N_0 = A_0 / \lambda = 5.35 \times 10^{15} \times 6000 = 32.1 \times 10^{18}$. Moles = $\frac{32.1 \times 10^{18}}{6 \times 10^{23}} = 5.35 \times 10^{-5} = 535 \times 10^{-7}$. Therefore, correct answer is 0535.