HOW TO SOLVE HALF-LIFE PROBLEMS EXPONENTIAL

What is the half life ?

Half-life is the time required for a quantity to reduce to half its initial value.

The term is used generally to characterize any type of exponential decay. When solving half-life problems, use the following formula:

A= A0 12tH

A = Amount remaining

A0 = Initial amount

t = time elapsed

H = half life

Problem 1 :

The half-life of carbon-14 is known to be 5720 years. If 300 grams of carbon-14 are stored for 1200 years, how many grams will remain?

Solution :

Initial amount A= 300 grams

Half life time (H) = 5720

Time elapsed (t) = 1200

A= A0 12tHA= 300 1212005720= 300 120.21= 300 (0.865)= 259.5= 260 grams

Approximately 260 grams will remain.

Problem 2 :

When Sophia drinks a brewed cup of coffee, she ingests 130 mg of caffeine into her system.

The half-life of caffeine in a typical adult is 5.5 hours. How much caffeine will be in her system 4 hours after she drinks the cup of coffee?

Solution :

Initial amount A= 130 mg

Half life time (H) = 4

Time elapsed (t) = 5.5

A= A0 12tHA= 130 1245.5= 130 120.73= 130(0.60)= 78 mg

Problem 3 :

A colony of bacteria decays so that the population days from now is given by

n(t) = 1000(1/2)t/4

a. What is the amount present when t = 0?

b. How much will be present in 4days?

c. What is the half-life?

Solution :

n(t) = 1000(1/2)t/4

a)  When t = 0

n(0) = 1000(1/2)0/4

= 1000(1)

= 1000

Present amount of bacteria is 1000.

b. When t = 4

n(4) = 1000(1/2)4/4

= 1000(1/2)1

= 1000(0.5)

= 500

c) n(t) = 1000(1/2)t/4

Initial amount = 1000, After half life time, it will become 500.

500 = 1000(1/2)t/4

1/2 = (1/2)t/4

(1/2)1(1/2)t/4

t/4 = 1

t = 4

So, after 4 years the amount of bacteria will become half.

Problem 4 :

The half-life of a radioactive isotope is 4 days. If 3.2 kg are present now, how much will be present after:

a. 4 days?

b. 8 days?

c. 20 days

d. t days?

Solution :

Initial amount A= 3.2 kg

Half life time (H) = 4

Time elapsed (t) = 

a. 4 days

A= A0 12tHA=3.2 1244=3.2 121=3.2(0.5)=1.6 kg

b. 8 days

A= A0 12tHA=3.2 1284=3.2 122=3.2(0.25)=0.8 kg

c. 20 days

A= A0 12tHA=3.2 12204=3.2 125=3.2(0.03125)=0.1 kg

d. t days

A= A0 12tHA=3.2 12t4

Problem 5 :

The half life of radium is about 1600 years. If 1 kg is present now, how much will be present after 

a)  3200 years   b)  16000 years     c)  800 years

Solution :

Initial amount A= 1 kg

Half life time (H) = 1600

Time elapsed (t) = in years

a) 3200 years

A= A0 12tHA=1 1232001600=122=0.25 kg

b) 16000 years

A= A0 12tHA=1 12160001600=1210=0.00098 kg

c)  800 years

A= A0 12tHA=1 128001600=1212=0.707 kg

Problem 6 :

When a certain medicine enters the bloodstream, it gradually dilutes, decreasing exponentially with a half-life of 3 days. The initial amount of the medicine in the bloodstream is A0 milliliters. What will the amount be 30 days later?

Solution :

Initial amount = A0

Half life time (H) = 3

Time elapsed (t) = 30 days

A= A0 12tHA= A012303= A0 1210= A0 (0.5)10= A0 (0.00097)= 0.097% of A0

Problem 7 :

A culture starts with 1500 bacteria and the number doubles every 30 minutes.

In the half-life formula, use a 2 instead of 1/2 to find a function that models the number of bacteria at time t 

b. Find the number of bacteria after 2 hours.

c. After how many minutes will there be 4000 bacteria?

Solution :

In 2 hours, we have 4 times 30 minutes

A= A0 12tH= 1500 (2)t30

b)

A= A0 12tH= 1500 (2)12030= 1500 (2)4= 24000

c)

A= A0 12tH=1500 (2)t30(2)t30(2)t30

After 42.50 minutes 4000 bacteria will be there.

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