InterviewSolution
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InterviewSolution
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(a) A radioactive sample has `6.0 xx 10^(18)` active nuclei at a certain instant. How many of these nuclei will still be the same active state after two half-lives? (b) Radioactive phosphours`-32` has a half-life of `14` days. A source containing this isotope has an initial activity of `10muCi`. (i) What is the activity of the source after `42` days? (ii) What time elapse before the activity of the source falls at `2.5muCi`? (c) `1g` of a radioactive substance disintergrates at the rate of `3.7 xx 10^(10) dps`. The atomis mass of the substance is `226`. Calculate its mean life. |
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Answer» (a) `N = N_(0)((1)/(2))^(n)` `n = 2` `N = (N_(0))/(4) = (6xx10^(18))/(4) = 15 xx 10^(17)` (b) `t = nT_(1//2)` `42 = nxx14 rArr n = 3` (i) `A = A_(0) ((1)/(2))^(n) = A_(0)((1)/(2))^(3) = (A_(0))/(8) = (10)/(8) muCi = 1.25muCi` (ii) `A = A_(0) ((1)/(2))^(n)` `2.5 = 10((1)/(2))^(n) rArr (1)/(4) = ((1)/(2))^(n)` `n =2` `t = nT_(1//2) = 2xx14 = 28` days (c) `N = (m)/(M) N_(A)` `A = lambdaN = (1)/(barT)(mN_(A))/(M)` `bar(T) = (mN_(A))/(MA) = (1xx6.023xx10^(23))/(226xx3.7xx10^(10)) sec` =` 7.2 xx 10^(10) sec` |
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