Question

One microgram of radioactive sodium $_{11}Na^{24}$ 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)

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