Radioactive
Decay and
Half-Life
The isotope Radium–228 undergoes beta decay as
shown in the following equation:
228
88
Ra 
228
89
Ac  e
0
1
The isotope Radium–228 undergoes beta decay as
shown in the following equation:
228
88
Ra 
228
89
As time goes on, the amount
of Radium that is still present
decreases, as the nuclei decay.
Ac  e
0
1
The isotope Radium–228 undergoes beta decay as
shown in the following equation:
228
88
Ra 
Start with 5.0 g
of radium–228
228
89
Ac  e
0
1
Decay of Radium–228
228
88
5.0
Mass (g)
4.0
3.0
2.0
1.0
0.0
0
5.8
11.6
17.4
23.2
Time (years)
29
34.8
40.6
46.4
Ra 
228
89
Ac  01 e
Decay of Radium–228
228
88
5.0
Mass
Mass (g)
4.0
3.0
2.0
1.0
0.0
0
5.8
11.6
17.4
23.2
Time (years)
29
34.8
40.6
46.4
Ra 
228
89
Ac  01 e
Decay of Radium–228
228
88
5.0
Mass (g)
4.0
3.0
2.0
1.0
0.0
0
5.8
11.6
17.4
23.2
Time (years)
29
34.8
40.6
46.4
Ra 
228
89
Ac  01 e
Decay of Radium–228
228
88
5.0
Mass (g)
4.0
3.0
2.0
1.0
0.0
0
t
5.8
11.6
17.4
23.2
Time (years)
29
34.8
40.6
46.4
Ra 
228
89
Ac  01 e
Decay of Radium–228
228
88
5.0
Mass (g)
4.0
3.0
2.0
1.0
0.0
0
t
5.8
11.6
17.4
23.2
Time (years)
29
34.8
40.6
46.4
Ra 
228
89
Ac  01 e
Decay of Radium–228
228
88
5.0
Mass (g)
4.0
3.4
3.0
2.0
1.0
0.0
0
t
5.8
11.6
17.4
23.2
Time (years)
29
34.8
40.6
46.4
Ra 
228
89
Ac  01 e
Decay of Radium–228
228
88
5.0
Mass (g)
4.0
3.4
3.0
2.0
1.0
0.0
0
t
5.8
11.6
17.4
23.2
Time (years)
29
34.8
40.6
46.4
Ra 
228
89
Ac  01 e
Decay of Radium–228
228
88
5.0
Mass (g)
4.0
3.0
2.0
1.0
Time “0”
0.0
0
5.8
11.6
17.4
23.2
Time (years)
29
34.8
40.6
46.4
Ra 
228
89
Ac  01 e
Decay of Radium–228
228
88
5.0
Mass (g)
4.0
3.0
2.0
1.0
Time “0”
0.0
0
5.8
11.6
17.4
23.2
Time (years)
29
34.8
40.6
46.4
Ra 
228
89
Ac  01 e
Decay of Radium–228
228
88
5.0
Mass (g)
4.0
3.0
The mass of radium–228
remaining at time 5.8 y = 2.5 g
2.0
1.0
0.0
0
5.8
11.6
17.4
23.2
Time (years)
29
34.8
40.6
46.4
Ra 
228
89
Ac  01 e
Decay of Radium–228
228
88
5.0
Mass (g)
4.0
3.0
The mass of radium–228
remaining at time 5.8 y = 2.5 g
2.5
2.0
1.0
0.0
0
5.8
11.6
17.4
23.2
Time (years)
29
34.8
40.6
46.4
Ra 
228
89
Ac  01 e
Decay of Radium–228
228
88
5.0
Mass (g)
4.0
3.0
The mass of radium–228 remaining
(2.5 g) is exactly one-half its initial
value of 5.0 g
2.5
2.0
1.0
0.0
0
5.8
11.6
17.4
23.2
Time (years)
29
34.8
40.6
46.4
Ra 
228
89
Ac  01 e
Decay of Radium–228
228
88
5.0
Mass (g)
4.0
3.0
The mass of radium–228 remaining
(2.5 g) is exactly one-half its initial
value of 5.0 g
2.5
2.0
1.0
0.0
0
5.8
11.6
17.4
23.2
Time (years)
29
34.8
40.6
46.4
Ra 
228
89
Ac  01 e
Decay of Radium–228
228
88
5.0
228
89
Ac  01 e
Half-Life
The time it takes for a radioactive isotope
to decay to half the mass it started with.
4.0
Mass (g)
Ra 
3.0
The mass of radium–228 remaining
(2.5 g) is exactly one-half its initial
value of 5.0 g
2.5
2.0
1.0
0.0
0
5.8
11.6
17.4
23.2
Time (years)
29
34.8
40.6
46.4
Decay of Radium–228
228
88
5.0
Mass (g)
4.0
3.0
2.5
2.0
1.0
0.0
0
5.8
11.6
17.4
23.2
Time (years)
29
34.8
40.6
46.4
Ra 
228
89
Ac  01 e
Decay of Radium–228
228
88
5.0
Ra 
Mass (g)
4.0
3.0
In the second 5.8 years, the mass
halved again from 2.5 g to 1.25 g
2.5
2.0
1.25
1.0
0.0
0
5.8
11.6
17.4
23.2
Time (years)
29
34.8
40.6
46.4
228
89
Ac  01 e
Decay of Radium–228
228
88
5.0
Ra 
228
89
Ac  01 e
Mass (g)
4.0
3.0
2.5
2.0
In the third 5.8 years, the mass
halved again from 1.25 g to 0.625 g
1.25
1.0
0.625
0.0
0
5.8
11.6
17.4
23.2
Time (years)
29
34.8
40.6
46.4
Decay of Radium–228
228
88
5.0
Ra 
228
89
Ac  01 e
Mass (g)
4.0
Every time one half-life passes, the mass of
the parent isotope drops to one-half of what
is was before that half-life.
3.0
2.5
2.0
1.25
1.0
0.625
0.0
0
5.8
11.6
17.4
23.2
Time (years)
29
34.8
40.6
46.4
Decay of Radium–228
228
88
5.0
Mass (g)
4.0
2.5 g
3.0
2.5
1.25 g
2.0
0.625 g
1.25
0.3125 g
1.0
0.625
0.0
0
1
5.8
2
11.6
3
17.4
4
23.2
Time (years)
Half-Lives
5
29
0.1563 g
6
34.8
7
40.6
8
46.4
Ra 
228
89
Ac  01 e
Decay of Radium–228
228
88
5.0
Mass (g)
4.0
3.0
2.5
2.0
1.25
1.0
0.625
0.0
0
1
5.8
2
11.6
3
17.4
4
23.2
Time (years)
Half-Lives
5
29
6
34.8
7
40.6
8
46.4
Ra 
228
89
Ac  01 e
Decay of Radium–228
228
88
5.0
Mass (g)
4.0
3.0
2.5
2.0
1.25
1.0
0.625
0.0
0
1
5.8
2
11.6
3
17.4
4
23.2
Time (years)
Half-Lives
5
29
6
34.8
7
40.6
8
46.4
Ra 
228
89
Ac  01 e
Decay of Radium–228
228
88
5.0
Mass (g)
4.0
3.0
2.5
2.0
1.25
1.0
0.625
0.0
0
5.8
11.6
17.4
23.2
Time (years)
29
34.8
40.6
46.4
Ra 
228
89
Ac  01 e
How many half-lives would pass
for the amount of a parent isotope
to drop to 30% of its initial mass?
How many half-lives would pass
for the amount of a parent isotope
to drop to 30% of its initial mass?
How many half-lives would pass
for the amount of a parent isotope
to drop to 30% of its initial mass?
How many half-lives would pass
for the amount of a parent isotope
to drop to 30% of its initial mass?
1.7
How many half-lives would pass
for the amount of a parent isotope
to drop to 30% of its initial mass?
Answer: 1.7 Half-lives
1.7
210
82
Pb 
210
83
Bi  01 e
210
82
Pb 
210
83
Bi  01 e
What is the half-life of lead-210?
210
82
Pb 
210
83
Bi  01 e
What is the half-life of lead-210?
210
82
Pb 
210
83
Bi  01 e
What is the half-life of lead-210?
210
82
Pb 
210
83
Bi  01 e
What is the half-life of lead-210?
210
82
Pb 
210
83
Bi  01 e
What is the half-life of lead-210?
210
82
Pb 
210
83
Bi  01 e
What is the half-life of lead-210?
Answer: 22 years
210
82
Pb 
210
83
Bi  01 e
What percent of the original
lead–210 remains after 80 years?
210
82
Pb 
210
83
Bi  01 e
What percent of the original
lead–210 remains after 80 years?
210
82
Pb 
210
83
Bi  01 e
What percent of the original
lead–210 remains after 80 years?
210
82
Pb 
210
83
Bi  01 e
What percent of the original
lead–210 remains after 80 years?
8%
210
82
Pb 
210
83
Bi  01 e
What percent of the original
lead–210 remains after 80 years?
Answer: 8%
8%
210
82
Pb 
210
83
What percentage of the total
possible bismuth-210 will be
produced after 60 years?
Bi  01 e
210
82
Pb 
210
83
What percentage of the total
possible bismuth-210 will be
produced after 60 years?
Bi  01 e
210
82
Pb 
210
83
What percentage of the total
possible bismuth-210 will be
produced after 60 years?
15%
Bi  01 e
210
82
Pb 
210
83
Bi  01 e
What percentage of the total
possible bismuth-210 will be
produced after 60 years?
15%
When 0% of the lead–210 remains, the
maximum total possible bismuth–210 will be
formed. So if 15% of the lead–210 remains,
that means (100-15) = 85% of the maximum
bismuth–210 will be formed.
210
82
Pb 
210
83
Bi  01 e
What percentage of the total
possible bismuth-210 will be
produced after 60 years?
15%
When 0% of the lead–210 remains, the
maximum total possible bismuth–210 will be
formed. So if 15% of the lead–210 remains,
that means (100-15) = 85% of the maximum
bismuth–210 will be formed.
210
82
Pb 
210
83
Bi  01 e
What percentage of the total
possible bismuth-210 will be
produced after 60 years?
15%
When 0% of the lead–210 remains, the
maximum total possible bismuth–210 will be
formed. So if 15% of the lead–210 remains,
that means (100-15) = 85% of the maximum
bismuth–210 will be formed.
210
82
Pb 
210
83
Bi  01 e
What percentage of the total
possible bismuth-210 will be
produced after 60 years?
15%
When 0% of the lead–210 remains, the
maximum total possible bismuth–210 will be
formed. So if 15% of the lead–210 remains,
that means (100-15) = 85% of the maximum
bismuth–210 will be formed.
210
82
Pb 
210
83
Bi  01 e
If one starts out with a 3.50 g sample of
lead–210, what mass of lead–210 will be
left after 45 years?
210
82
Pb 
210
83
Bi  01 e
If one starts out with a 3.50 g sample of
lead–210, what mass of lead–210 will be
left after 45 years?
210
82
Pb 
210
83
Bi  01 e
If one starts out with a 3.50 g sample of
lead–210, what mass of lead–210 will be
left after 45 years?
210
82
Pb 
210
83
Bi  01 e
If one starts out with a 3.50 g sample of
lead–210, what mass of lead–210 will be
left after 45 years?
210
82
Pb 
210
83
Bi  01 e
If one starts out with a 3.50 g sample of
lead–210, what mass of lead–210 will be
left after 45 years?
24%
210
82
Pb 
210
83
Bi  01 e
If one starts out with a 3.50 g sample of
lead–210, what mass of lead–210 will be
left after 45 years?
24%
24% of 3.50 g
 0.24  3.50 g
 0.84 g
210
82
Pb 
210
83
Bi  01 e
If one starts out with a 3.50 g sample of
lead–210, what mass of lead–210 will be
left after 45 years?
24%
24% of 3.50 g
 0.24  3.50 g
 0.84 g
210
82
Pb 
210
83
Bi  01 e
If one starts out with a 3.50 g sample of
lead–210, what mass of lead–210 will be
left after 45 years?
24%
24% of 3.50 g
 0.24  3.50 g
 0.84 g
210
82
Pb 
210
83
Bi  01 e
If one starts out with a 3.50 g sample of
lead–210, what mass of lead–210 will be
left after 45 years? Ans: 0.84 g
24%
24% of 3.50 g
 0.24  3.50 g
 0.84 g
Using Equations
to Do
Radioactivity
Problems
99
42
The half-life of the isotope Mo is 66.0 hours.
An initial sample contains 2.00 grams of molybdenum–99.
What mass of molybdenum–99 will be left in the sample
after 8 days and 6 hours?
99
42
The half-life of the isotope Mo is 66.0 hours.
An initial sample contains 2.00 grams of molybdenum–99.
What mass of molybdenum–99 will be left in the sample
after 8 days and 6 hours?

24 h 
 8 d  1 d   6 h  192 h  6 h  198 h


99
42
The half-life of the isotope Mo is 66.0 hours.
An initial sample contains 2.00 grams of molybdenum–99.
What mass of molybdenum–99 will be left in the sample
after 8 days and 6 hours?

24 h 
 8 d  1d   6 h  192 h  6 h  198 h


99
42
The half-life of the isotope Mo is 66.0 hours.
An initial sample contains 2.00 grams of molybdenum–99.
What mass of molybdenum–99 will be left in the sample
after 8 days and 6 hours?

24 h 
 8 d  1 d   6 h  192 h  6 h  198 h


99
42
The half-life of the isotope Mo is 66.0 hours.
An initial sample contains 2.00 grams of molybdenum–99.
What mass of molybdenum–99 will be left in the sample
after 8 days and 6 hours?

24 h 
 8 d  1d   6 h  192 h  6 h  198 h


99
42
The half-life of the isotope Mo is 66.0 hours.
An initial sample contains 2.00 grams of molybdenum–99.
What mass of molybdenum–99 will be left in the sample
after 8 days and 6 hours?

24 h 
 8 d  1 d   6 h  192 h  6 h  198 h


99
42
The half-life of the isotope Mo is 66.0 hours.
An initial sample contains 2.00 grams of molybdenum–99.
What mass of molybdenum–99 will be left in the sample
after 8 days and 6 hours?

24 h 
 8 d  1 d   6 h  192 h  6 h  198 h


99
42
The half-life of the isotope Mo is 66.0 hours.
An initial sample contains 2.00 grams of molybdenum–99.
What mass of molybdenum–99 will be left in the sample
after 8 days and 6 hours?

24 h 
 8 d  1 d   6 h  192 h  6 h  198 h


99
42
The half-life of the isotope Mo is 66.0 hours.
An initial sample contains 2.00 grams of molybdenum–99.
What mass of molybdenum–99 will be left in the sample
after 8 days and 6 hours?

24 h 
 8 d  1 d   6 h  192 h  6 h  198 h


198 h
 3 half-lives
66 h/half-life
99
42
The half-life of the isotope Mo is 66.0 hours.
An initial sample contains 2.00 grams of molybdenum–99.
What mass of molybdenum–99 will be left in the sample
after 8 days and 6 hours?

24 h 
 8 d  1 d   6 h  192 h  6 h  198 h


198 h
 3 half-lives
66 h/half-life
99
42
The half-life of the isotope Mo is 66.0 hours.
An initial sample contains 2.00 grams of molybdenum–99.
What mass of molybdenum–99 will be left in the sample
after 8 days and 6 hours? (3 half-lives)
99
42
The half-life of the isotope Mo is 66.0 hours.
An initial sample contains 2.00 grams of molybdenum–99.
What mass of molybdenum–99 will be left in the sample
after 8 days and 6 hours? (3 half-lives)
1 1 1
2.00 g       0.25 g
 2 2 2
99
42
The half-life of the isotope Mo is 66.0 hours.
An initial sample contains 2.00 grams of molybdenum–99.
What mass of molybdenum–99 will be left in the sample
after 8 days and 6 hours? (3 half-lives)
1 1 1
2.00 g       0.25 g
 2 2 2
Multiply by ½ for
each half-life.
99
42
The half-life of the isotope Mo is 66.0 hours.
An initial sample contains 2.00 grams of molybdenum–99.
What mass of molybdenum–99 will be left in the sample
after 8 days and 6 hours? (3 half-lives)
1 1 1
2.00 g       0.25 g
 2 2 2
99
42
The half-life of the isotope Mo is 66.0 hours.
An initial sample contains 2.00 grams of molybdenum–99.
What mass of molybdenum–99 will be left in the sample
after 8 days and 6 hours? (3 half-lives) Answer: 0.25 g
1 1 1
2.00 g       0.25 g
 2 2 2
An Equation We Can Use:
An Equation We Can Use:
massremaining
1
 massinitial   
 2
n
An Equation We Can Use:
massremaining
1
 massinitial   
 2
n
Where n = the # of half-lives that have passed
Here’s an example question:
Here’s an example question:
The radioisotope thorium–234 has a half-life of
24 days.
Here’s an example question:
The radioisotope thorium–234 has a half-life of
24 days. A sample initially contains 5.000 g of
thorium–234.
Here’s an example question:
The radioisotope thorium–234 has a half-life of
24 days. A sample initially contains 5.000 g of
thorium–234. What mass of thorium–234 will be left in
the sample after 120 days?
The radioisotope thorium–234 has a half-life of 24 days. A sample initially
contains 5.000 g of thorium–234. What mass of thorium–234 will be left in
the sample after 120 days?
Number of Half-lives =
massremaining
120 days
= 5 half-lives
24 days/half-life
1
 massinitial   
 2
5
n
1
= 5.000 g ×  
 2
 1 
= 5.000 g ×   = 0.156 g
 32 
The radioisotope thorium–234 has a half-life of 24 days. A sample initially
contains 5.000 g of thorium–234. What mass of thorium–234 will be left in
the sample after 120 days?
Number of Half-lives =
massremaining
120 days
= 5 half-lives
24 days/half-life
1
 massinitial   
 2
5
n
1
= 5.000 g ×  
 2
 1 
= 5.000 g ×   = 0.156 g
 32 
The radioisotope thorium–234 has a half-life of 24 days. A sample initially
contains 5.000 g of thorium–234. What mass of thorium–234 will be left in
the sample after 120 days?
Number of Half-lives =
massremaining
120 days
= 5 half-lives
24 days/half-life
1
 massinitial   
 2
5
n
1
= 5.000 g ×  
 2
 1 
= 5.000 g ×   = 0.156 g
 32 
The radioisotope thorium–234 has a half-life of 24 days. A sample initially
contains 5.000 g of thorium–234. What mass of thorium–234 will be left in
the sample after 120 days?
Number of Half-lives =
massremaining
120 days
= 5 half-lives
24 days/half-life
1
 massinitial   
 2
5
n
1
= 5.000 g ×  
 2
 1 
= 5.000 g ×   = 0.156 g
 32 
The radioisotope thorium–234 has a half-life of 24 days. A sample initially
contains 5.000 g of thorium–234. What mass of thorium–234 will be left in
the sample after 120 days?
Number of Half-lives =
massremaining
120 days
= 5 half-lives
24 days/half-life
1
 massinitial   
 2
5
n
1
= 5.000 g ×  
 2
 1 
= 5.000 g ×   = 0.156 g
 32 
The radioisotope thorium–234 has a half-life of 24 days. A sample initially
contains 5.000 g of thorium–234. What mass of thorium–234 will be left in
the sample after 120 days? Answer: 0.156 g
Number of Half-lives =
massremaining
120 days
= 5 half-lives
24 days/half-life
1
 massinitial   
 2
5
n
1
= 5.000 g ×  
 2
 1 
= 5.000 g ×   = 0.156 g
 32 
The radioisotope thorium–234 has a half-life of 24 days. A sample initially
contains 5.000 g of thorium–234. What mass of thorium–234 will be left in
the sample after 120 days? Answer: 0.156 g
Number of Half-lives =
massremaining
120 days
= 5 half-lives
24 days/half-life
1
 massinitial   
 2
5
n
1
= 5.000 g ×  
 2
 1 
= 5.000 g ×   = 0.156 g
 32 
The radioisotope thorium–234 has a half-life of 24 days. A sample initially
contains 5.000 g of thorium–234. What mass of thorium–234 will be left in
the sample after 120 days? Answer: 0.156 g
Number of Half-lives =
massremaining
120 days
= 5 half-lives
24 days/half-life
1
 massinitial   
 2
5
n
1
= 5.000 g ×  
 2
 1 
= 5.000 g ×   = 0.156 g
 32 