Hi,

 

Question:

A radioactive substance has a half-life of 20 days.

 

i)

How much time is required so that only 1/32 of the initial amount remains?

ii)

Find the rate of decay at this time.

 

My Answer:

i)

Let "A" represent the initial amount.

Let "t" represent the time in days.

 

(1/32)A = A(.5)^(t/20)

 

(1/32) = (.5)^(t/20)

 

log_.5(1/32) = t/20

 

5 = t/20

 

t = 100 days

ii)

Let "A" represent the initial amount.

Let "t" represent the time in days.

f(t) = A(.5)^(t/20)

 

f ' (t) = A (ln(.5)) ((.5)^(t/20)) (1/20)

 

f ' (100) = A (ln(.5)) ((.5)^(100/20)) (1/20)

= A (ln(.5)) ((.5)^(5)) (1/20)

= A (-0.001083042)

 

The rate of decay at 100 days is equivilant to [ -A(0.001083042) ].

 

 

 

 

Are these answers correct?

1/32 requires 5 half-lives, so 100 days is correct. The rate of decay looks correct as well, though one might question the sign, since it's already known the rate is negative.

These questions are simple compared to the questions in the preceeding unit of the course... I just wanted to make sure it wasn't a trick :/

 

 

 

Thanks!