Question

The half-life of radium-226 is 1620 yr. Given a sample of 1 g of radium-226, the quantity left Q(t) (in g) after t years is given by:

Q(t)= 1/2^t/1620

Required:
a. Convert this to an exponential function using base e.
b. Verify that the original function and the result from part (a) yield the same result for Q(0), Q(1620), and Q(3240).

Answer 1

(a)
`e^(-0.000428 t)`

Explanation:

We are given that

Half life of radium-226=1620 yr

The quantity left Q(t) after t years is given by

`Q(t)=((1)/(2))^{(t)/(1620)}`

a. We have to convert the given function into an exponential function using base e.

`Q(t)=((1)/(2))^{(t)/(1620)}`

=
`(((1)/(2))^t)^{(1)/(1620)`

=
`e^(ln(1/2) t/1620)`

`=e^({(ln(1/2))/(1620)t)`

=
`e^(-0.000428 t)`

(b)

`Q(0)=e^(-0.000428 * 0)`

=1

From original function

Q(0)=1

`Q(1620)=((1)/(2))^{(t)/(1620)}`

`Q(1620)=(1)/(2)=0.5`

From exponential function

`Q(1620)=e^(-0.000428 * 1620)`

`=0.499\approx 0.5`

`Q(3240)=((1)/(2))^{(3240)/(1620))=0.25`

`Q(3240)=e^(-0.000428 * 3240)`

Q(3240)=0.249=
`\approx 0.25`

Hence, verified.