In the formula;
A = I ⋅ eᵏᵗ
A is the amount of radioactive material remaining from an initial amount I at a given time t and k is a negative constant determined by the nature of the material. A certain radioactive isotope has a half-life of approximately 1000 years.
How many years would be required for a given amount of this isotope to decay to 45% of that amount?

Respuesta :

Answer:

1152

Step-by-step explanation:

given that in the formula

[tex]A = I (e^{kt)}[/tex]

half life is 1000 years

In other words when t =1000 A=l/2

[tex]A = I (e^{kt)}[/tex]

Since A = l/2 we get

[tex]0.5=e^{1000k} \\1000k = ln 0.5 = -0.6931\\k = -0.00069[/tex]

So equation is

[tex]A= le^{-0.00069t}[/tex]

When A/l = 0.45, we have

[tex]0.45=e^{-0.00069t} \\-0.00069t = -0.79851\\t=1152[/tex]

i.e. in 1152 years we expect to decay to 45%.

The age of the radioactive isotope can be determined from the amount

of the isotope remaining and the rate it decomposes.

Correct response:

  • The number of years required is 1,152 years

Method used for the calculation of the time duration required

The function for the rate of decay for the radioactive isotope is; A = [tex]I \cdot e ^{k \cdot t}[/tex]

Where;

A = The amount of material left after time t

k = The rate

t = The time duration

I = The initial amount of the material at t = 0

The half life of the given isotope = 1,000

Therefore;

When t = 1,000, A = [tex]\dfrac{I}{2} [/tex], which gives;

[tex]\dfrac{I}{2} = \mathbf{I \cdot e^{1000 \cdot k}}[/tex]

[tex]\dfrac{1}{2} = e^{1000 \cdot k}[/tex]

[tex]\mathbf{ln\left(\dfrac{1}{2} \right)} = 1000 \cdot k[/tex]

[tex]k = \dfrac{ln\left(\frac{1}{2}\right) }{1000} \approx \mathbf{ -6.9315 \times 10^{-4}}[/tex]

When the isotope decays to 45%, we have;

[tex]45\% = 0.45 = e^{\left(-6.9315 \times 10^{-4} \right)\times t}[/tex]

Which gives;

-6.9315 × 10⁻⁴ × t = ㏑(0.45)

[tex]t \approx \mathbf{ \dfrac{ln(0.45)}{-6.9315 \times 10^{-4}}} \approx 1,152[/tex]

  • The number of years required for the given amount of isotope to decay to 45% is t1,152 years

Learn more about radioactive disintegration here:

https://brainly.com/question/20518385

https://brainly.com/question/13446873

RELAXING NOICE
Relax