A green plant contains radioactive C $$14$$ and C $$12$$ isotope. If in a sample of dead old wood of this plant $$2.5$$×$$10-6$$ $$mol14C$$ and $$7.5$$×$$10-6$$ $$mol14$$ N are present then calculate value of Z. Assuming all ‘N’ is obtained due to disintegration of $$14C$$.Where $$Z=Age$$ of woodtHalf life of $$14C$$ $$isotopet1/2$$

Question

A green plant contains radioactive C  $$14$$ and C  $$12$$ isotope. If in a sample of dead old wood of this plant $$2.5$$×$$10-6$$ $$mol14C$$ and $$7.5$$×$$10-6$$ $$mol14$$ N are present then calculate value of Z. Assuming all ‘N’ is obtained due to disintegration of  $$14C$$.Where $$Z=Age$$ of woodtHalf life of $$14C$$ $$isotopet1/2$$

Infinity Learn · Accepted Answer

$$614C$$→  $$714N+70e$$∧$$t=ln$$ $$N0Nt=ln10$$×$$10$$−$$62.5$$×$$10$$−$$6=ln$$ $$4ln$$ $$2t1/2t=2$$ ln $$2t=2$$ $$t1/2z=2$$

Kinetics of nuclear reactions or nuclear chemistry

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$_{6}^{14} C {\to_{}^{}}_{7}^{14} N +_{7}^{0} e$
$\land t = l n \frac{N_{0}}{N_{t}} = l n (\frac{10 \times 10^{- 6}}{2.5 \times 10^{- 6}}) = l n 4$
$\frac{l n 2}{t_{1 / 2}} t = 2 l n 2$
$t = 2 t_{1 / 2}$
$z = 2$

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