The half life of radium-226 is 1620 years.Calculate (a)the decay constant (b)the time it takes 60% of a given sample to decay, and (c)the initial activity of 1gm of pure radium-226


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Question Number 41361 by Necxx last updated on 06/Aug/18

$T h e h a l f l i f e o f r a d i u m - 226 i s 1620$ $y e a r s . C a l c u l a t e (a) t h e d e c a y c o n s t a n t$ $(b) t h e t i m e i t t a k e s 60 % o f a g i v e n$ $s a m p l e t o d e c a y, a n d (c) t h e i n i t i a l$ $a c t i v i t y o f 1 g m o f p u r e r a d i u m - 226$
	Answered by MJS last updated on 06/Aug/18

	$initial amount = A_{0}$ $amount after t years = A_{t}$ $A_{t} = A_{0} {(\frac{1}{2})}^{\frac{t}{1620}}$ $A_{t} = A_{0} e^{- λ t}$ $e^{- λ t} = {(\frac{1}{2})}^{\frac{t}{1620}}$ $- λ t = \frac{t}{1620} \ln \frac{1}{2}$ $λ = \frac{\ln 2}{1620} \approx 4.2787 \times 10^{- 4}$ $A_{0} = 100 %$ $40 = 100 {(\frac{1}{2})}^{\frac{t}{1620}}$ $\frac{2}{5} = {(\frac{1}{2})}^{\frac{t}{1620}}$ $\ln \frac{2}{5} = \frac{t}{1620} \ln \frac{1}{2}$ $t = 1620 \frac{\ln \frac{2}{5}}{\ln \frac{1}{2}} \approx 2141.5 years$
	Commented by Necxx last updated on 06/Aug/18

	$T h e s t r e n g t h o f a r a d i o a c t i v e s a m p l e$ $m a y b e s p e c i f i e d a t a g i v e n t i m e b y$ $i t s a c t i v i t y a n d t h e u n i t c o m m o n l y$ $e m p l o y e d i s t h e C u r i e (C i) a n d i s$ $d e f i n e d a s :$ $1 C i = 3.70 \times 10^{10} d e c a y / s$
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