In a radioactive sample, 19 40 K nuclei either decay into stable 20 40 C a nuclei with decay constant 4.5 × 10 - 10 per year or into stable 18 40 A r nuclei with decay constant 0.5 × 10 - 10 per year. Given that in this sample all the stable 20 40 C a and 18 40 A r nuclei are produced by the 19 40…

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In a radioactive sample, 19 40 K nuclei either decay into stable 20 40 C a nuclei with decay constant 4.5 × 10 - 10 per year or into stable 18 40 A r nuclei with decay constant 0.5 × 10 - 10 per year. Given that in this sample all the stable 20 40 C a and 18 40 A r nuclei are produced by the 19 40 K nuclei only. In time t × 10 9 years, if the ratio of the sum of stable 20 40 C a and 18 40 A r nuclei to the radioactive 19 40 K nuclei is 99 , the value of t will be : [Given l n 10 = 2.3 ]

Aicharya · Accepted Answer

Given that, 19 49 K decays in two stable 20 40 C a o f 18 40 A r Nuclei. Let there be N 0 active nuclei of 19 40 K present at t = 0 and none of 20 40 C a and 18 40 A r present of t = 0 . Where λ 1 (decay constant for K → C a ) = 4.5 × 10 - 10 y e a r and λ 2 (decay constant for K → A r ) = 0.5 × 10 - 10 y e a r These form two parallel nuclear reactions. ∴ Equivalent decay consent, λ = λ 1 + λ 2 = 5 × 10 - 10 y e a r Now, number of active nuclei of 19 40 K left at t = t = N And according to Law of Radioactivity, N = N 0 e - λ t ........(i) Now, according to question, N 0 - N N = 99 ⇒ N 0 = 100 N ⇒ N = N 0 100 Then using equation (i) N 0 100 = N 0 e - λ t ⇒ l n 100 = λ t ⇒ t = 2 l n 10 λ = 2 × 2.3 5 × 10 - 10 = 9.2 × 10 9 y e a r s

In a radioactive sample, 19 40 K nuclei either decay into stable 20 40 C a nuclei with decay constant 4.5 ×…

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