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Radioactive substance disintegrate by 10%in 1 month .How much fraction will disintegrate in 4 months?

1 Answer

May 14, 2018

$35.6 %$ $35.6%$ decayed after 4 months

Explanation:

We have the equation:
$N = N_{0} e^{- λ t}$ $N=N_0e^(-lambdat)$ , where:

$N$ $N$ = current number of radioactive nuclei remaining
$N_{0}$ $N_0$ = starting number of radioactive nuclei remaining
$t$ $t$ = time passed ( $s$ $s$ though can be hours, days, etc.)
$λ$ $lambda$ = decay constant $(\frac{ln (2)}{t_{\frac{1}{2}}})$ $(ln(2)/t_(1/2))$ ( $s^{- 1}$ $s^-1$ , though in the equation uses the same unit of time as $t$ $t$ )

$10 %$ $10%$ decay, so $90 %$ $90%$ remain

$0.9 N_{0} = N_{0} e^{- λ}$ $0.9N_0=N_0e^(-lambda)$ ( $t$ $t$ being taken in months, and $l a, b d a$ $la,bda$ being ${month}^{- 1}$ $"month"^-1$ )

$λ = - ln (0.9) = 0.11 {month}^{- 1}$ $lambda=-ln(0.9)=0.11"month"^-1$ (to 2 d.p)

$a N_{0} = N_{0} e^{- 0.11 (4)}$ $aN_0=N_0e^(-0.11(4))$

$100 % a = 100 % - (e^{- 0.11 (4)} \cdot 100 %) = 100 % - 64.4 % = 35.6 %$ $100%a=100%-(e^(-0.11(4))*100%)=100%-64.4%=35.6%$ decayed

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