Question

2.The number of grams A of a certain radioactive substance present at time, in yearsfrom the present, t is given by the formulaA = 45e^-0.0045ta. What is initial amount of this substance?b. What is half-life of this substance?c. How much will be around in 2500 years?

Answer 1

Answer:

Given that,

The number of grams A of a certain radioactive substance present at time, in years

from the present, t is given by the formula

$A=45e^{-0.0045(t)}$

a) To find the initial amount of this substance

At t=0, we get

$A=45e^{-0.0045(0)}$$A=45e^0$

We know that e^0=1 ( anything to the power zero is 1)

we get,

$A=45$

The initial amount of the substance is 45 grams

b)To find thehalf-life of this substance

To find t when the substance becames half the amount.

A=45/2

Substitute this we get,

$\frac{45}{2}=45e^{-0.0045(t)}$

$\frac{1}{2}=e^{-0.0045(t)}$

Taking natural logarithm on both sides we get,

$\ln (\frac{1}{2})=-0.0045(t)^{}$$(-1)\ln (\frac{1}{2})=0.0045(t)$$\ln (\frac{1}{2})^{-1}=0.0045(t)$$\ln (2)=0.0045(t)$$0.6931=0.0045(t)$$t=\frac{0.6931}{0.0045}$$t=154.02$

Half-life of this substance is 154.02

c) To find the amount of substance will be present around in 2500 years

Put t=2500

we get,

$A=45e^{-0.0045(2500)}$$A=45e^{-11.25}$$A=45\times0.000013=0.000585$$A=0.000585$

The amount of substance will be present around in 2500 years is 0.000585 grams