Question

AIATS -7 Practice Test (Code A)
21. In an ore at the time of estimation, the ratio of U238 to.
Pb206 is 1:7. Calculate the age of ore, assuming that
all the lead present in the ore is the final stable product
of U238. (Half life of U238 = 4.5 * 10 years)
(1) 1.87* 109 yr 2 13.5 * 109 yr
(3) 8.5 x 109 yr (4) 9* 10º yr​

Answer 1

Answer:

Total age of the given ore is

$t = 13.5 \times 10^9 yr$

Explanation:

If the ratio of U238 and Pb206 in the given ore is 1 : 7

so we can say that the fraction of U present at that time is

$U = \frac{1}{8}$

so it is three times the half life

so we will have

$t = 3 T_{1/2}$

given that

$T_{1/2} = 4.5 \times 10^9 yr$

so we will have

$t = 3(4.5 \times 10^9)$

$t = 13.5 \times 10^9 yr$

#Learn

Topic : Half life

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