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Question: In nature a decay chain series starts with \({}_{90}^{232}Th\) and finally terminates at \({}_{82}^{...

In nature a decay chain series starts with 90232Th{}_{90}^{232}Th and finally terminates at 82208Pb.{}_{82}^{208}Pb. A thorium ore sample was found to

contain 8×10−5mL8 \times 10^{- 5}mL of helium at STP and 5×10−7g5 \times 10^{- 7}g of

232Th.232Th. Find the age of the ore sample assuming the source of helium to be only decay of 232Th232Th. Also assume complete retention of helium within the ore. (Half life of 232Th=1.39×1010232Th = 1.39 \times 10^{10}years)

A

6.89×1096.89 \times 10^{9}years

B

4.89×1094.89 \times 10^{9} years

C

3.69×1093.69 \times 10^{9}years

D

6.893×10106.893 \times 10^{10} years

Answer

4.89×1094.89 \times 10^{9} years

Explanation

Solution

No. of moles of helium=8×10−522400= \frac{8 \times 10^{- 5}}{22400}

90232Th→82208Pb+624He{}_{90}^{232}Th \rightarrow_{82}^{208}Pb + 6_{2}^{4}He

No. of 90232Th{}_{90}^{232}Th moles which have disintegrated =8×10−56×22400= \frac{8 \times 10^{- 5}}{6 \times 22400}

Mass of 90232Th{}_{90}^{232}Th which have disintegrated

=8×10−5×2326×22400=1.3809×10−7g= \frac{8 \times 10^{- 5} \times 232}{6 \times 22400} = 1.3809 \times 10^{- 7}g

Mass of 232Th232Th left, ′Nt′=5×10−7g'N_{t}' = 5 \times 10^{- 7}g

′N0′=(5×10−7+1.3809×10−7)=6.3809×10−7g'N_{0}' = (5 \times 10^{- 7} + 1.3809 \times 10^{- 7}) = 6.3809 \times 10^{- 7}g

Applying t=2.303λlog⁡N0Nt=2.3030.693×1.39×1010log⁡6.3809×10−75×10−7=4.89×109t = \frac{2.303}{\lambda}\log\frac{N_{0}}{N_{t}} = \frac{2.303}{0.693} \times 1.39 \times 10^{10}\log\frac{6.3809 \times 10^{- 7}}{5 \times 10^{- 7}} = 4.89 \times 10^{9} years