There are 0.618 μ g of 206 Pb and 0.238 μ g of 238 U in a rock. If T50 of 238 U is 1.5 × 109 yr, age of the rock is

Question

There are 0.618 μ g of 206 Pb and 0.238 μ g of 238 U in a rock. If T50 of 238 U is 1.5 × 109 yr, age of the rock is (A) 1.5 × 109 yr (B) 3.0 × 109 yr (C) 4.5 × 109 yr (D) 0.75 × 109 yr

Tardigrade Admin · Accepted Answer

Based on rock-dating (1+([ Pb ]/[ U ])=(2)n (1+((0.618 × 10-6/206)/(0.238 × 10-6)238))=(2)n (2)2=(2)n n=2 2=(t/T50) t=2 × T50=3 × 109 yr

Q. There are $0.618 μg$ of $^{206} P b$ and $0.238 μg$ of $^{238} U$ in a rock. If $T_{50}$ of $^{238} U$ is $1.5 \times 1 0^{9} yr$ , age of the rock is

$1.5 \times 1 0^{9} yr$

$3.0 \times 1 0^{9} yr$

$4.5 \times 1 0^{9} yr$

$0.75 \times 1 0^{9} yr$

Based on rock-dating
$(1 + \frac{[ P b ]}{[ U ]} = (2)^{n}$
$(1 + \frac{\frac{0.618 \times 1 0 ^{- 6}}{206}}{\frac{0.238 \times 1 0 ^{- 6}}{238}}) = (2)^{n}$
$(2)^{2} = (2)^{n}$
$n = 2$
$2 = \frac{t}{T _{50}}$
$t = 2 \times T_{50} = 3 \times 1 0^{9} yr$

Q. There are 0.618μg of 206Pb and 0.238μg of 238U in a rock. If T50​ of 238U is 1.5×109yr, age of the rock is

1.5×109yr

3.0×109yr

4.5×109yr

0.75×109yr