Question

At radioactive equilibrium, the ratio between $2$ atoms of radioactive elements $A$ and $B$ is $3\times {10}^9:1$. If $t_{1/2}$ of $A$ is $10$ yrs, what is $t_{1/2}$ of $B$?

• A. $30$ yrs
• B. $3$ yrs
• C. $3.3$ yrs
• D. None of these

Answer 1

Answer: (C)

$3.3$ yrs

$k_a=\frac{0.693}{t_{1/2}^A}=\frac{0.693}{10}{yr}^{-1}=0.0693{year}^{-1}$
We know $\frac{A}{B}=3\times {10}^9$
We find that
$k_B=0.33$(using the given info)
So, $(from{k}_B=\frac{0.693}{t_{1/2}^B})t_{\frac{1}{2}}=\frac{0.693}{0.33}\sim 3.3years.\left(optionC\right)$