In a radioactive sample. `._(19)^(40)K` nuclei either decay into stable `._(20)^(40)Ca` nuclei with decay constant `4.5 xx 10^(-10)` per year or into stable `._(18)^(40)Ar` nuclei with decay constant `0.5 xx 10^(-10)` per year. Given that in this sample all the stable `._(20)^(40)Ca and ._(18)^(40)Ar` nuclei are produced by the `._(19)^(40)K` nuclei only. In time `t xx 10^9` years. If the ratio of the sum of stable `._(20)^(40)Ca` and `._(18)^(40)Ar` nuclei to the radioactive `._(19)^(40)K` nuclei is 99. The value of t will be. [Given : In 10 = 2.3]

Options

A. 9.2
B. 1.15
C. 0.02
D. 0.01

Answer

A

Explanation

This is an example of parallel radioactive decay and effective decay constant can be calculated by adding individual decay constants.`rArr lambda= lambda_(1) + lambda_(2) = 5 xx 10^(-10)` per year. If `N_(0)` is the initial number of radioactive nuclei then number of radioactive nuclei at some instant t can be written as follows:
`N=N_(0)e^(-lambdat)`............(i)
Number of stable nuclei: `N_("stable")= N_(0)-N`
Number of radioactive nuclei: `N_("radioactive")=N`
Ratio of the stable to the radioactive nuclei is given to be 99. Hence we can write the following.
`(N_(0)-N)/N = 99 rArr N_(0)/N -1 =99 rArr N_(0)/N =100`
Using equation (i) we can write:
`N_(0)/N =e^(lambdat)=100 rArr lambdat=ln100 rArr lambdat=2 ln10`
`rArr lambdat=4.6 rArr t = 4.6/(5 xx 10^(-10)) =9.2 xx 10^(9)` years.