Question

Half-life for radioactive $^{14}C$ is 5760 yr. In how many years, 200 mg of $^{14}C$ will be reduced to 25 mg?

• A. 5760 yr
• B. 11520 yr
• C. 17280 yr
• D. 23040 yr

Answer 1

The Correct Option is C

Half-life of radioactive
$t_{1/2}2 5760 yr, [R]_0 = 200 mg$
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, $[R] = 25 mg
Rate constant (k) =$\frac{0.6932}{t_{1/2}}=\frac{0.6932}{5760}yr^{-1}$
$\therefore\, \, \, \, \, \, \, \, \, \, \, \, k=\frac{2.303}{t}log \frac{| R |_0}{| R |}$
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, (R_0$ = 200 mg inital amount)
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \frac{0.6932}{t}=\frac{2.303}{t}log\frac{200}{25}$
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, $(R = 25 mg reduced amount)
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =\frac{2.303}{t}$ log 8
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \frac{0.6932}{5760}=\frac{2.303}{t}\times 0.9030$
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, t=\frac{5760 \times 2.303 \times 0.9030}{0.6932} $
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =\frac{11978.54}{0.6932}=17280 yr$