Respuesta :
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
[tex]\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}[/tex]
To find value of k
[tex]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}[/tex]
Percentage of radioactive element is:
Amount after 900 years[tex]=\frac{P(900)}{P_{o}}\times 100\\\\=\frac{0.68P_{o}}{P_{o}}\times 100\%\\\\=68\%[/tex]
