If A(t) is the activity after t years have elapsed, and if A0 is the original activity:

– A(0) = A0

– A(90) = A0/2

-A(180) = A(90)/2 = A0/4

..

-A(n*90) = A((n-1)*90)/2 = .. = (A0 / 2^(n-1) ) / 2 = A0 / 2^n

Here, n is an integer, but we can interpolate with a real argument: A(t) = A0 / 2^(t/90).

To get the answer, we use logarithm to solve A(t1) = A0 * 0.05.

ln A0 + ln 0.05 = ln A0 – (t/90) * ln 2

t = – 90 * ln(0.05) / ln 2 ~ 389 years.

So:

In 90 years – it’s activity would be 50%

In 180 years – it’s activity would be 25% (half of 50)

Etc.

It can be shown that, for exponential decay, the half-life t1 / 2 obeys this relation:

t1/2 = (ln2)/λ

where ln(2) is the natural logarithm of 2 (approximately 0.693), and

λ is the decay constant, a positive constant used to describe the rate of exponential decay.

The half-life is related to the mean lifetime τ by the following relation:

t1/2 = (ln2) X τ

Look at the link (below) to fill in the values:

[ln(0.5)]/k=90

k= -7.7

we arent given any amount but we know we want 0.05 of the initial amount.

0.05A=Ae^-7.7t

0.05=e^-7.7t

ln0.05=-7.7lne

ln0.05=-7.7t

ln0.05/-7.7=t

t=0.389

time units will be what you were told in the question.

thats what i got, so a bit out to your answer?