In a sample of rock, the ratio of 206Pb to 238U nuclei is found to be 0.5. The age of the rock is (Given half life of U238decay to =4.5×109 year and answer if initially the rock had equal nuclei of 206Pb and 238U)

NPbNUr=12 NUrNPb=2 NUrN0=23 23=12n 32=2n ln(32)=n ln 2 ln(32)=tT ln 2 t=Tln(3/2)ln(2)⇒t=4.5×109ℓn(3/2)ℓn(2)

In a sample of rock, the ratio of 206Pb to 238U nuclei is found to be 0.5. The age of the rock is (Given half life of U238decay to =4.5×109 year and answer if initially the rock had equal nuclei of 206Pb and 238U)

NPbNUr=12 NUrNPb=2 NUrN0=23 23=12n 32=2n ln(32)=n ln 2 ln(32)=tT ln 2 t=Tln(3/2)ln(2)⇒t=4.5×109ℓn(3/2)ℓn(2)

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In a sample of rock, the ratio of $^{206} Pb$ to $^{238} U$ nuclei is found to be 0.5. The age of the rock is (Given half life of $U^{238}$ decay to $= 4.5 \times 10^{9}$ year and answer if initially the rock had equal nuclei of $^{206} Pb$ and $^{238} U$ )

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$4.5 \times 10^{9} l n 3 year$

$2.25 \times 10^{9} year$

$4.5 \times 10^{9} \frac{l n (\frac{3}{2})}{l n 2} year$

$2.25 \times 10^{9} l n (\frac{3}{2}) year$

answer is C.

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$\frac{N_{Pb}}{N_{Ur}} = \frac{1}{2} \frac{N_{Ur}}{N_{Pb}} = 2 \frac{N_{Ur}}{N_{0}} = \frac{2}{3} \frac{2}{3} = \frac{1}{2^{n}} \frac{3}{2} = 2^{n} l n (\frac{3}{2}) = n l n 2 l n (\frac{3}{2}) = \frac{t}{T} l n 2 t = T \frac{l n (3 / 2)}{l n (2)}$

$\Rightarrow t = 4.5 \times 10^{9} \frac{ℓn (3 / 2)}{ℓn (2)}$

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In a sample of rock, the ratio of 206Pb to 238U nuclei is found to be 0.5. The age of the rock is (Given half life of U238decay to =4.5×109 year and answer if initially the rock had equal nuclei of 206Pb and 238U)

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In a sample of rock, the ratio of $^{206} Pb$ to $^{238} U$ nuclei is found to be 0.5. The age of the rock is (Given half life of $U^{238}$ decay to $= 4.5 \times 10^{9}$ year and answer if initially the rock had equal nuclei of $^{206} Pb$ and $^{238} U$ )