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WBJEEIn a radioactive sample, 1940K nuclei either decay into stable 2040Ca nuclei with...
In a radioactive sample, 4019K nuclei either decay into stable 4020Ca nuclei with decay constant 4.5×10-10 per year or into stable 4018Ar nuclei with decay constant 0.5×10-10 per year. Given that in this sample all the stable 4020Ca and 4018Ar nuclei are produced by the 4019K nuclei only. In time t×109 years, if the ratio of the sum of stable 4020Ca and 4018Ar nuclei to the radioactive 4019K nuclei is 99, the value of t will be : [Given ln 10=2.3 ]
Solution:
Given that, 4919K decays in two stable 4020Ca of 4018Ar Nuclei. Let there be N0 active nuclei of 4019K present at t=0 and none of 4020Ca and 4018Ar present of t=0 .
Where λ1 (decay constant for K→Ca ) =4.5×10-10year and
λ2 (decay constant for K→Ar ) =0.5×10-10year
These form two parallel nuclear reactions.
∴ Equivalent decay consent, λ=λ1+λ2=5×10-10year
Now, number of active nuclei of 4019K left at (t=t)=N
And according to Law of Radioactivity,
N=N0e-λt ........(i)
Now, according to question,
N0-NN=99⇒N0=100N⇒N=N0100
Then using equation (i)
N0100=N0e-λt
⇒ln 100=λt⇒t=2ln10λ=2×2.35×10-10=9.2×1010 years
Where λ1 (decay constant for K→Ca ) =4.5×10-10year and
λ2 (decay constant for K→Ar ) =0.5×10-10year
These form two parallel nuclear reactions.
∴ Equivalent decay consent, λ=λ1+λ2=5×10-10year
Now, number of active nuclei of 4019K left at (t=t)=N
And according to Law of Radioactivity,
N=N0e-λt ........(i)
Now, according to question,
N0-NN=99⇒N0=100N⇒N=N0100
Then using equation (i)
N0100=N0e-λt
⇒ln 100=λt⇒t=2ln10λ=2×2.35×10-10=9.2×1010 years
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