There are 0-618 mu g of -206-Pb and 0-238mu g of -238-U in a rock- If T-50- of -238-U is 1-5 times 10-9 yr- age of the rock is -1-5 times 10-9 yr3-0 times 10-9 yr0-75 times 10-9 yr4-5 times 10-9 yr

Question

Toppr Admin · Accepted Answer

-Pb-0-618-xD7-10-x2212-6206mol-3-xD7-10-x2212-9mol

There are $0.618 μ g$ of $^{206} P b$ and $0.238 μ g$ of $^{238} U$ in a rock. If $T_{50}$ of $^{238} U$ is $1.5 \times 10^{9} y r$ , age of the rock is :
$1.5 \times 10^{9} y r$
$3.0 \times 10^{9} y r$
$0.75 \times 10^{9} y r$
$4.5 \times 10^{9} y r$

$[P b] = \frac{0.618 \times 10^{- 6}}{206} m o l = 3 \times 10^{- 9} m o l$

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 :1.5×109yr3.0×109yr0.75×109yr4.5×109yr

[Pb]=0.618×10−6206mol=3×10−9mol

There are $0.618 μ g$ of $^{206} P b$ and $0.238 μ g$ of $^{238} U$ in a rock. If $T_{50}$ of $^{238} U$ is $1.5 \times 10^{9} y r$ , age of the rock is :
$1.5 \times 10^{9} y r$
$3.0 \times 10^{9} y r$
$0.75 \times 10^{9} y r$
$4.5 \times 10^{9} y r$

$[P b] = \frac{0.618 \times 10^{- 6}}{206} m o l = 3 \times 10^{- 9} m o l$