Rate of Decay

[sysD]

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?

[swansont]

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.

[sysD]

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!