The amount of radioactive lead remaining after 3 years is 545gm.
The amount of radioactive lead remaining after 8 years is 464gm.
The amount of radioactive lead remaining after 15 years is 371gm.
The half-tie for radio active lead is 21.66 years.
Given that,
A sample of 600 g of radioactive lead-210 decays to polonium-210,
According to the function A(t) = 600e ^-0.032t,
Where, t is time in years.
We have to find,
Find the amount of radioactive lead remaining after.
According to the question,
Function [tex]A(t) = 600e^{(-0.032t)}[/tex]
[tex]A(3) = 600\times e^{(-0.032t)} \\\\A (3) = 600 \times \dfrac{1}{e^{0.032\times 3} } \\\\e^{0.096} = 1.10075\\\\A (3) = 600 \times \dfrac{1}{e^{0.096} } \\\\A (3) = 600 \times \dfrac{1}{1.10075} } \\\\A(3) = \dfrac{600}{1.10075}\\\\A(3) = 545gm[/tex]
The amount of radioactive lead remaining after 3 years is 545gm.
[tex]A(8) = 600\times e^{(-0.032t)} \\\\A (8) = 600 \times \dfrac{1}{e^{0.032\times 8} } \\\\e^{0.0256} = 1.129175\\\\A (8) = 600 \times \dfrac{1}{e^{0.0256} } \\\\A (8) = 600 \times \dfrac{1}{1.129175} } \\\\A(8) = \dfrac{600}{1.129175}\\\\A(8) = 464gm[/tex]
The amount of radioactive lead remaining after 8 years is 464gm.
[tex]A(15) = 600\times e^{(-0.032t)} \\\\A (15) = 600 \times \dfrac{1}{e^{0.032\times 15} } \\\\e^{0.48} = 1.61607\\\\A (8) = 600 \times \dfrac{1}{e^{0.48} } \\\\A (8) = 600 \times \dfrac{1}{1.61607} } \\\\A(8) = \dfrac{600}{1.61607}\\\\A(8) = 371gm[/tex]
The amount of radioactive lead remaining after 15 years is 371gm.
[tex]A(15) = 600\times \dfrac{1}{2} = 600e^{-0.032t} \\\\= e^{(-0.032t)} = \dfrac{1}{2}\\\\Taking \ log \ on \ both \ sides\\\\= -0.032t = ln2\\\\= t = \dfrac{ln2}{-0.032}\\\\= t = 21.660\\\\= t = 21.66 years[/tex]
The half-tie for radio active lead is 21.66 years.
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