Question

One microgram of radioactive sodium ${}_{11}N{a}^{24}$ with a half-life of 15 h was injected into a living system for a bio-assay. How long will it take for the radioactivity to fall to 25% of the initial value?

• A. 60 h
• B. 22.5 h
• C. 375 h
• D. 30 h

Answer 1

Answer: (D)

$t_{1/2}o{f}_{11}N{a}^{24}=15h$
$[R{]}_0=1.0\mu g$
$[R]=1.0\times \frac{25}{100}=\frac{25}{100}=0.25\mu g$
$k=\frac{0.6932}{t_{1/2}}=\frac{0.6932}{15}{h}^{-1}$
$\therefore k=\frac{2.303}{t}log\frac{[R{]}_0}{[R]}$
$\therefore \frac{0.6932}{15}=\frac{2.303}{t}log\frac{1}{0.25}$
$\frac{0.6932}{15}=\frac{2.303}{t}log\frac{100}{25}=\frac{2.303}{t}log$4
$\frac{0.6932}{15}=\frac{2.303}{t}\times 0.6020$
$t=\frac{15\times 2.303\times 0.6020}{0.6932}$
$=\frac{20.7995}{0.6932}=30.00h$