22huffmansridn
Answered

Discover a world of knowledge and community-driven answers at IDNLearn.com today. Our platform is designed to provide accurate and comprehensive answers to any questions you may have.

NEED HELP ASAP!!!!

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

We value your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. Thank you for choosing IDNLearn.com. We’re committed to providing accurate answers, so visit us again soon.