The document describes a lab that originally had 230g of a radioactive substance called Z with a half-life of 8 years. It provides the decay formula and shows calculations to determine: 1) The rate of decay is -0.08664 2) It will take about 16 years for the amount to decay to 57.5g 3) It will take about 28 years for the amount to decay to 20g 4) The lab would have originally had 1000g about 17 years ago

1. Solutions

2. A lab has about 230g of a Radio Active
Substance, quot;Zquot;.The Half­Life of Z is 8
years.
rt
Formula: A=Pe

3. A) Find the rate at which it will decay at:
Formula to find the half life we
1 r8 know to get 1/2 it will take 8 years
= e
2 so all that is need is to find r
ln 1 = r8 By taking the ln of both side we
2 get ln(e)^r8 which will leave r8
ln 1 because ln(e) is log base e of e
r = 2
that so (e)^r8 is an exponent that
8
on base e will give you r8
r = ­0.08664

4. B) How long until there is 57.5g left? 20g?
A=57.5
57.5 = 230e ­0.08664t Once you know the rate all you need
to do is Find A for these questions
1
= e ­0.08664t Isolate e^­0.08664t by dividing
4 by 230.
Like the previous question take the
1
ln = ­0.08664t ln of both sides to easily get rid of
4
the e.
ln 1
t = Isolate the t and solve.
4
­0.08664 In about 16 years 1/4 or 57.5g
t = 16.0006 will remain.

5. A=20
­0.08664t Same steps as previous
20 = 230e question
2
= e ­0.08664t
23
ln 2 = ­0.08664t
23
ln 2
t = 23
­0.08664 In about 28 years an a few
t = 28.1896 months 1/4 or 57.5g will remain.

6. C) If the lab originally had 1000g of Z, how long
ago was there that much of it?
1000 = 230e ­0.08664t C is the same as B, but working with
negtaive time, which means it has
1000
= e ­0.08664t occured some time in the past.
230
ln 1000= ­0.08664t
230
ln 1000
t = 230
­0.08664 In about 17 years ago there was
t = ­16.9630 1000g of z in the lab