4.4.2 applications of first oder ODE (radioactive decay)

A particular radioactive substance has a half-life of 20 days. After 50 days, only 100 mg has remained.
Find the amount of the radioactive substance initially.

= 0, = 0 Initial P Integrating both sides, because the
half-life radioactive has
1
= decayed
= +
= 20 , 1
= 2 0

= 50 , = 100

= +
From =

Separating variables t and P, = ∙ 
=
1
=

MOOC MAT438/ UiTM

A particular radioactive substance has a half-life of 20 days. After 50 days, only 100 mg has remained.
Find the amount of the radioactive substance initially.

= 0, = 0 to find C To find k, substitute = 20 , = 1 0 into 
to find k 2
1
= 20 , = 2 0 1 = 0 20
2 0

= 50 , = 100 20 = 0.5 half-life

= 

To find C, substitute = 0, = 0 into  20 = 0.5 decayed
from = = −0.03465 (into )
0 = 0 1
→ = 0 −0.03465 
= 0 (into )

→ = 0 

MOOC MAT438/ UiTM

A particular radioactive substance has a half-life of 20 days. After 50 days, only 100 mg has remained.
Find the amount of the radioactive substance initially.

= 0, = 0 to find C 0 = 565.48 mg
to find k
1 Into  the amount of the radioactive
= 20 , = 2 0 substance initially is 565.48 mg

= 50 , = 100

An expressions for the population at any time t,

= 0 −0.03465 

To find 0, substitute = 50 , = 100 into  0.03465 50
100
100 = 0 −0.03465 50 100 = 0 − 0.03465 × 50
0 = 100 0.03465 ×50 100

0 = − 0.03465 × 50

MOOC MAT438/ UiTM



Extract the information

(t and P)

The isotope 253Es (einsteinium-253) was used to produced mendelevium (atomic number 101). 253Es decays at

a rate proportional to the amount present. After 12 days, two-thirds of the isotope remains. Determine its

half-life. Initial P is

not given

= 0, = 0 Initial P Integrating both sides,

2 two-thirds 1
= 12 , = 3 0 remains =
= +
=? , 1 half-life
= 2 0
න = +

= + 
From = = ∙
=
Separating variables t and P,

1
=

MOOC MAT438/ UiTM

The isotope253Es (einsteinium-253) was used to produced mendelevium (atomic number 101). 253Es decays at
a rate proportional to the amount present. After 12 days, two-thirds of the isotope remains. Determine its
half-life.

= 0, = 0 to find C → = 0 

= 12 , 2 to find k To find k, substitute = 12 , = 2 0 into 
= 3 0 3

= ? , 1 2 0 = 0 12
= 2 0 3

=  12 = 2
3

To find C, substitute = 0, = 0 into  12 = 2
from = 3
0 = 0 1
= 0 into  = −0.03378 into

→ = 0 −0.03378 decayed


MOOC MAT438/ UiTM

The isotope253Es (einsteinium-253) was used to produced mendelevium (atomic number 101). 253Es decays at
a rate proportional to the amount present. After 12 days, two-thirds of the isotope remains. Determine its
half-life.

= 0, = 0 to find C To find half-life, substitute = 1 0 into 
2

= 12 , 2 to find k 1 0 = 0 −0.03378
= 3 0 2

1 −0.03378 = 1 Half-life
= 2 0 2
= ? , Into 

−0.03378 = 0.5

An expressions for the population at any time t, 0.5 Half-life = 1
= −0.03378 2



= 0 −0.03378 

= 20.5 days

the half-life of einsteinium-253 is 20.5 days

MOOC MAT438/ UiTM

MOOC MAT438/ UiTM

Extract the information
(t and P)

Carbon-14 has a half-life of 5,715 years. A fossil has lost 75% of its original amount of C-14. How old is the
fossil? How much C-14 remains after 50,000 years?

Initial P is

not given

= 0, = 0 Initial P Integrating both sides,

1 1
= 5,715 , = 2 0 half-life න = + =
i) = ? , = 0.25 0 Lost 75% = +

ii) = H50ow,00ol0d , =? 25% remains = +
= ∙
Amount

remains
From =

Separating variables t and P, = 

1
=

MOOC MAT438/ UiTM

Carbon-14 has a half-life of 5,715 years. A fossil has lost 75% of its original amount of C-14. How old is the
fossil? How much C-14 remains after 50,000 years?

= 0, = 0 to find C to find k = 0 into  
1 → = 0

= 5,715 , = 2 0 To find k, substitute = 5,715 , = 1 0 into 
i) = ? , = 0.25 0 2
1
ii) = 50,000 , =? 2 0 = 0 5715

5715 = 1 half-life
2
= 

To find C, substitute = 0, = 0 into  5715 = 1 decayed
from =
0 = 0 1 2

into 0.5
= 5715 = −0.00012

→ 0.5 

= 0 5715

MOOC MAT438/ UiTM

Carbon-14 has a half-life of 5,715 years. A fossil has lost 75% of its original amount of C-14. How old is the
fossil? How much C-14 remains after 50,000 years?

= 0, = 0 to find C 0.5 = 0.25
1
to find k 5715
= 5,715 , = 2 0

i) = ? , = 0.25 0 Into  How old is 0.5
ii) = 50,000 , =? the fossil? 5715 = 0.25

5715
= 0.25 ∙ 0.5

An expressions for the population at any time t, = 11,430 years

0.5 

= 0 5715

i) Substitute = 0.25 0 into  the fossil is 11,430 years old

0.25 0 0.5

= 0 5715

MOOC MAT438/ UiTM

Carbon-14 has a half-life of 5,715 years. A fossil has lost 75% of its original amount of C-14. How old is the
fossil? How much C-14 remains after 50,000 years?

= 0, = 0 to find C ii) Substitute = 50,000 into 
1
to find k 50,0 0 0 0.5
= 5,715 , = 2 0
= 0 5715
i) = ? , = 0.25 0 ( ℎ (50000 × (0.5)))/5715

ii) = 50,000 , =? Into  = 0.0023 0 since the original amount
= 0.0023 × 100% 0 was not given, write the
An expressions for the population at any time t, = 0.23% 0
answer in percentage
0.5 form

= 0 5715 

0.5  After 50,000 years, C-14 remains 0.23% from
the original amount
= 0 5715

MOOC MAT438/ UiTM

MOOC MAT438/ UiTM