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Q.A piece of wood shows C14 activity which is 20% of the activity found today. If the decay follows first order kinetics, calculate the age of the wood sample. (Given t½ for ¹⁴₆C = 5770 years.)

Goa GbshseGBSHSE Class 12 Board Exam 2018Subjective· 2mImportance★★★★★
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Using first-order radioactive decay kinetics with the given half-life, the wood's age works out to about 1.34 × 10⁴ years.

Radioactive decay follows first-order kinetics, so:

k=0.693t1/2=0.6935770 yr=1.201×10−4 yr−1k = \frac{0.693}{t_{1/2}} = \frac{0.693}{5770\ \text{yr}} = 1.201\times10^{-4}\ \text{yr}^{-1}

The first-order integrated rate law (in terms of activity, which is proportional to the remaining amount of 14C^{14}C):

t=2.303klog⁡A0At = \frac{2.303}{k}\log\frac{A_0}{A}

Here the current activity AA is 20% of the original activity A0A_0, so A0A=10020=5\dfrac{A_0}{A} = \dfrac{100}{20} = 5.

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