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JEE Advanced · Physics · 28. Nuclear Physics

In a radioactive sample, 1940K nuclei either decay into stable 2040Ca nuclei with decay constant 4.5×10-10 per year or into stable 1840Ar nuclei with decay constant 0.5×10-10 per year. Given that in this sample all the stable 2040Ca and 1840Ar nuclei are produced by the 1940K nuclei only. In time t×109 years, if the ratio of the sum of stable 2040Ca and 1840Ar nuclei to the radioactive 1940K nuclei is 99, the value of t will be : [Given ln 10=2.3 ]

  1. A 1.15
  2. B 2.3
  3. C 4.6
  4. D 9.2
Verified Solution

Answer & Solution

Correct Answer

(D) 9.2

Step-by-step Solution

Detailed explanation

Given that, 1949K decays in two stable 2040Ca of 1840Ar Nuclei. Let there be N0 active nuclei of 1940K present at t=0 and none of 2040Ca and 1840Ar present of t=0 .
Where λ1 (decay constant for KCa ) =4.5×10-10year and
λ2 (decay constant for KAr ) =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 1940K left at t=t=N
And according to Law of Radioactivity,
N=N0e-λt ........(i)
Now, according to question,
N0-NN=99N0=100NN=N0100
Then using equation (i)
N0100=N0e-λt
\(\Rightarrow \ln 100=\lambda t \Rightarrow t=\frac{2 \ln 10}{\lambda}=\frac{2 \times 2.3}{5 \times 10^{-10}}\) \(=9.2 \times 10^{10}\) years
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