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Standard XII

Mathematics

Intersection

A sample of ...

Question

A sample of $^{238} U$ (half-life $= 4.5$ $\times 10^{9}$ years) ore is found to contain 23.8 g of $238 U$ and 20.6 g of $^{206} P b$ . Calculate the age of the ore.

t = 4.5 \times 10^{9}

years

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t = 4.4 \times 10^{9}

years

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t = 5.5 \times 10^{9}

years

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None of these

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Solution

The correct option is A $t = 4.5 \times 10^{9}$ years
$(1 + \frac{[P b]}{[U]}) = 2^{y}$
$(1 + \frac{20.6 / 206}{23.8 / 238}) = 2^{y}$
$2 = 2^{y}$
$y = 1$
$\frac{t}{t_{50}} = 1$
$t = 4.5 \times 10^{9}$ years

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A sample of 238U (half-life =4.5×109 years) ore is found to contain 23.8 g of 238U and 20.6 g of 206Pb. Calculate the age of the ore.

The correct option is A t=4.5×109 years(1+[Pb][U])=2y(1+20.6/20623.8/238)=2y2=2yy=1tt50=1t=4.5×109 years

A sample of $^{238} U$ (half-life $= 4.5$ $\times 10^{9}$ years) ore is found to contain 23.8 g of $238 U$ and 20.6 g of $^{206} P b$ . Calculate the age of the ore.

The correct option is A $t = 4.5 \times 10^{9}$ years
$(1 + \frac{[P b]}{[U]}) = 2^{y}$
$(1 + \frac{20.6 / 206}{23.8 / 238}) = 2^{y}$
$2 = 2^{y}$
$y = 1$
$\frac{t}{t_{50}} = 1$
$t = 4.5 \times 10^{9}$ years