The half-life for the $\alpha$-decay of uranium $\ _{92}U^{238}$is $4.47 imes 10^{9}\ $yr. If a rock contains sixty percent of its original $\ _{92}U^{238}$atoms, its age is $\lbrack\log{6 = 0.778;\ \log{2 = 0.3 brack}}$

Question

The half-life for the $\alpha$-decay of uranium $\ _{92}U^{238}$is $4.47 	imes 10^{9}\ $yr. If a rock contains sixty percent of its original $\ _{92}U^{238}$atoms, its age is $\lbrack\log{6 = 0.778;\ \log{2 = 0.3brack}}$

Dr. Abubacker · Accepted Answer

(a) Here, $T = 4.47 	imes 10^{9}yr$
$\frac{N}{N_{0}} = \frac{60}{100} = \left( \frac{1}{2} ight)^{n}or\ 2^{n} = \frac{10}{6}$
$n\log{2 = \log{10 - \log{6 = 1 - 0.778 = 0.222}}}$
$n = \frac{0.222}{\log 2} = \frac{0.222}{0.3} = 0.74$
$t = nT = 0.74 	imes 4.47 	imes 10^{9}yr$
$= 3.3 	imes 10^{9}yr$

Q. The half-life for the $\alpha$-decay of uranium $\ _{92}U^{238}$is $4.47 \times 10^{9}\ $yr. If a rock contains sixty percent of its original $\ _{92}U^{238}$atoms, its age is $\lbrack\log{6 = 0.778;\ \log{2 = 0.3\rbrack}}$

Explanation

Reviewed by Dr. Abubacker

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