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Q.

The half life for the $α$ 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 $[\log 6 = 0.778; \log 2 = 0.3]$

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a

$3.3 \times 10^{9} Y$

b

$6.6 \times 10^{9} Y$

c

$1.2 \times 10^{8} Y$

d

$5.4 \times 10^{7} Y$

answer is A.

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Detailed Solution

Here, $T = 4.47 \times 10^{9} y r$

$\frac{N}{N_{0}} = \frac{60}{100} = {(\frac{1}{2})}^{n} o r 2^{n} = \frac{10}{6}$

$n \log 2 = \log 10 - \log 6 = 1 - 0.778 = 0.222$

$n = \frac{0.222}{\log 2} = \frac{0.222}{0.3} = 0.74$

$t = n T = 0.74 \times 4.47 \times 10^{9} y r$

$= 3.3 \times 10^{9} y r$

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The half life for the α decay of uranium 92u238 is 4.47×109 yr. if a rock contains sixty percent of its original 92u238 atoms, its age is [log6=0.778;log2=0.3]