Question

Radioactive Decay

Answer 1

Answer:

Percentage of (226Ra) after 900 years is 68%

Step-by-step explanation:

Let P(t) be the amount of (226Ra) present at any time t

Half life of (226 Ra) = 1599 years

If P₀ is initial amount of (226 Ra) then after 1599 years

P(1599)=P₀/2

Decay i amount of radioactive substance is related to time t as

$\frac{dP}{dt}=kP(t)\\\\\frac{1}{P}\,dP=kdt\\\\Integrating\,\, both\,\,sides\\\\ln|P|=kt+c\\\\P(t)=Ce^{kt}\\\\at \,\, t=0\,\, P(0)=P_{o}\\\\P(0)=Ce^{k0}\\\\P_{o}=C\\\\then\\\\P(t)=P_{o}e^{kt}$

To find value of k

$at\,\, t=1599\,years\\\\P(1599)=\frac{P_{o}}{2}\\\\then\\\\\frac{P_{o}}{2} =P_{o}e^{k(1599)}\\\\\frac{1}{2} =e^{k(1599)}\\\\ln|\frac{1}{2}|=k(1599)\\\\k=\frac{ln|\frac{1}{2}|}{1599}=-4.3\times 10^{-3}\\\\\implies P(t)=P_{o}e^{-4.3\times 10^{-3}t}\\\\at\,\, t=900 \\\\P(900)=P_{o}e^{-4.3\times 10^{-3}(900)}\\\\P(900)=0.68P_{o}$

Percentage of radioactive element is:

Amount after 900 years$=\frac{P(900)}{P_{o}}\times 100\\\\=\frac{0.68P_{o}}{P_{o}}\times 100\%\\\\=68\%$