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InterviewSolution
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How many atoms of `0.1 g`-atom of a radioacitve isotope `._(Z)X^(A)` (half life = 5 days) will decay during the 11th day? |
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Answer» `N_(0) = 0.1g` atoms, `t = 10` day and `t_(1//2) = 5` day `lambda = (2.303)/(t) log_(10) (N_(0))/(N)` `(0.6963)/(5) = (2.303)/(10) log_(10) (0.1)/(N)` `:. N_(10)` i.e., amount left after 10 day `0.0250g` atom Similarly, if `t=11day` `(0.693)/(5)=(2.303)/(11)log__(10)(0.1)/(N)` `:. N_(11)` i.e., amount left after 11 day `= 0.0218g` atom `:.` Amount decayed in `11th` day `= N_(10) - N_(11)` `= 0.0250-0.0218 = 3.2xx10^(-3)g` atoms `= 3.2xx6.023xx10^(23) xx 10^(-3)` atoms `= 1.93xx10^(21)` atoms |
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