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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. Find the amount of radioactive lead remaining after
​(a) 3 yr
(b) 8 yr
(c) 15 yr
(d) find the half-tie

round to the nearest integer as needed.

Sagot :

600g - 210d + -0.032= 15yr or c

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]

  • After 3 years the amount of radioactive lead,

[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.

  • After 8 years the amount of radioactive lead,

[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.

  • After 15 years the amount of radioactive lead,

[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.

  • The half-tie for radioactive lead,

[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.

To know more about Radioactivity click the link given below.

https://brainly.com/question/12520966

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