1 mole = 6.02 x 1023 atoms
1u = 1.66 x 10−27kg
1. Americium-241 has a decay constant of 5.07 x 10−11s−1. What is the activity of 1 mole of americium-241?

A
=
λ
N
=
5.07
×

10

−
11

×
6.02
×

10

23

=
3.05
×

10

13

Bq

{\displaystyle A=\lambda N=5.07\times 10^{-11}\times 6.02\times 10^{23}=3.05\times 10^{13}{\mbox{ Bq}}}

This is why we only need very small samples for use in, for example, smoke detectors. In fact, safety considerations necessitate a small sample - otherwise we would all have cancer!
2. How many g of lead-212 (λ = 18.2μs−1) are required to create an activity of 0.8 x 1018Bq?

N
=

A
λ

=

0.8
×

10

18

18.2
×

10

−
6

=
4.40
×

10

22

nuclei

=
4.40
×

10

22

×
212
×
1.66
×

10

−
27

=
15.5

g

{\displaystyle N={\frac {A}{\lambda }}={\frac {0.8\times 10^{18}}{18.2\times 10^{-6}}}=4.40\times 10^{22}{\mbox{ nuclei}}=4.40\times 10^{22}\times 212\times 1.66\times 10^{-27}=15.5{\mbox{ g}}}

3. How long does it take for 2 kg of lead-212 to decay to 1.5 kg of lead-212?
Mass is proportional to the number of atoms, so:

m
=

m

0

e

−
λ
t

{\displaystyle m=m_{0}e^{-\lambda t}}

m

m

0

=

e

−
λ
t

{\displaystyle {\frac {m}{m_{0}}}=e^{-\lambda t}}

m

0

m

=

e

λ
t

{\displaystyle {\frac {m_{0}}{m}}=e^{\lambda t}}

λ
t
=
ln
⁡

m

0

m

{\displaystyle \lambda t=\ln {\frac {m_{0}}{m}}}

t
=

ln
⁡

m

0

m

λ

=

ln
⁡

2
1.5

18.2
×

10

−
6

=
15807

s

=
4.39

hours

{\displaystyle t={\frac {\ln {\frac {m_{0}}{m}}}{\lambda }}={\frac {\ln {\frac {2}{1.5}}}{18.2\times 10^{-6}}}=15807{\mbox{ s}}=4.39{\mbox{ hours}}}

4. Where does the missing 0.5 kg go?
It becomes another isotope - in this case, mercury-208. Some also becomes alpha particles.
5. Some americium-241 has an activity of 3kBq. What is its activity after 10 years?

A
=

A

0

e

−
λ
t

=
3000
×

e

−
5.07
×

10

−
11

×
10
×
365.24
×
24
×
60
×
60

=
2.952

kBq

{\displaystyle A=A_{0}e^{-\lambda t}=3000\times e^{-5.07\times 10^{-11}\times 10\times 365.24\times 24\times 60\times 60}=2.952{\mbox{ kBq}}}

6. This model of radioactive decay is similar to taking some dice, rolling them once per. second, and removing the dice which roll a one or a two. What is the decay constant of the dice?
The decay constant is the probability of removing a die - λ = ⅓.
7. If you started out with 10 dice, how many dice would you have left after 10s? What is the problem with this model of radioactive decay?

N
=

N

0

e

−
λ
t

=
10
×

e

−

1
3

×
10

=
0.357

dice

{\displaystyle N=N_{0}e^{-\lambda t}=10\times e^{-{\frac {1}{3}}\times 10}=0.357{\mbox{ dice}}}

Obviously, you can't have 0.357 dice. The problem with this model of radioactive decay is that, once you have sufficiently few nuclei, the decay ceases to be continuous. As time passes, the pattern becomes relatively more random. The model also says that the number of nuclei will always decrease. In reality, since there can only be an integer number of nuclei, there will eventually come a point when there are no nuclei left.