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Chemistry
Half-life for radioactive 14C is 5760 yr. In how many years, 200 mg of 14C will be reduced to 25 mg?
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Q. Half-life for radioactive $^{14}C$ is 5760 yr. In how many years, 200 mg of $^{14}C$ will be reduced to 25 mg?
AIPMT
AIPMT 1995
A
5760 yr
12%
B
11520 yr
15%
C
17280 yr
64%
D
23040 yr
10%
Solution:
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} $
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =\frac{11978.54}{0.6932}=17280 yr$