# Half-life Problems And Answers Examples

## Half-life Problems And Answers Examples

### Problem 1

In 24 days, a radioactive isotope decreased in mass from 64 grams to 2 grams. What is the half-life of the radioactive material?

The half-life of the radioactive material is 4.8 days.

#### Explanation

We are going to apply two methods to arrive at our answer

##### Method 1: Conventional method

64 g to 32 g = 1 half-life

32 g to 16 g = 2 half-life

16 g to 8 g = 3 half-life

8 g to 4 g = 4 half-life

4 g to 2 g = 5 half-life

If 5 half-life is equal to 24 days. Then 1 half-life will be T

Therefore, T x 5 half-life = 1 half-life x 24 days

Hence

T = (1 half-life x 24 days) / 5 half-life = 4.8 days

##### Method 2: Zhepwo Method

Decay time, t = 24 days

Half-life, T = ?

Initial mass, N1 = 64 g

Final mass, N2 = 2 g

We will use the formula

R = N1 / N2 = 64 / 2 = 32

T = t / (log2R) = 24 / (log232) = 24 / 5 = 4.8 days

Therefore, the half-life of the radioactive material is 4.8 days

### Problem 2

The half-life of radioactive material is 6 hours. What quantity of 1 kg of the material would decay in 24 hours?

The final answer to the above question is (15/16) 0.9375 kg

#### Explanation

We will also apply two methods to solve the above question

##### Method 1: Conventional Method

After 6 hours, 1/2 kg decays, and 1/2 kg will remain

Another 6 hours, 1/2 of 1/2 kg decays, 1/4 kg remains

6 hours after, 1/2 of 1/4 kg decays, 1/8 kg remains

After another 6 hours, 1/2 of 1/8 kg decays, 1/16 kg remains

Therefore, the material decayed would be:

1 kg – 1/16 kg = (15 / 16) kg = 0.9375 kg

##### Method 2: Zhepwo Method

Half-life, T = 6 hrs

Initial mass, N1 = 1 kg

Mass of the decayed material, Nd = ?

Formula

T = t/n

This implies that

n = t / T = 24 / 6 = 4

R = 2n = 24 = 16

Nd = N1 (R – 1 / R) = 1 (16-1 / 16) = (15/16) kg

Therefore, the quantity of 1 kg of the material that would decay in 24 hours is 15/16 kilograms or 0.9375 kilograms.

### Problem 3

A radioactive sample initially contains N atoms. After three (3) half-lives, what is the number of atoms that have disintegrated?

The final answer to the above question is (7/8) N

#### Explanation

We will use two methods to solve the above question

##### Method 1: Conventional method

First step: After 1 half-life, N/2 decay, N/2 remain

Second Step: After 2 half-life, 1/2 of N/2 decay, N/4 remain

Third step: After 3 half-life, 1/2 of N/4 decay, N/8 remains

The number of atoms disintegrated = Sum of the disintegrated atoms or fractions

This is now equal to

= N/2 + (1/2 x N/2) + (1/2 x N/4) = N/2 + N/4 + N/8 = (7/8)N

##### Method 2: Zhepwo Method

Alternatively, we can also apply the following steps to solve the problem using Zhepwo method:

Initial number of atoms, N1 = N

Number of half-lives, n = 3

Also, the number of atoms decayed, Nd = ?

Solution

R = 2n = 23 = 8

Nd = N1 (R – 1 / R) = N ( 8 – 1 / 8) = N x 7/8 = (7/8)N

Therefore, the number of atoms that have disintegrated is (7/8)N

### Problem 4

After three half-lives, the fraction of a radioactive material that has decayed is

The final answer to the above question is 7/8

#### Explanation

We will use two methods

##### Method 1: Conventional Method

1 half-life implies 1/2 decays, ,1/2 remains

2 half-life shows that 1/4 decays, 1/4 remains

3 half-life, 1/8 decays, 1/8 remains

Hence, the fraction decayed = original fraction – remaining fraction = (1 – 1/8) = 7/8

Note that the original fraction is 1/1 which is equal to 1

##### Method 2: Zhepwo Method

Number of half-lives, n = 3

The fraction decayed fd = ?

Disintegration ratio, R = 2n = 23 = 8

fd = R – 1 / R = 8 – 1 / 8 = 7/8

Therefore, the fraction of radioactive material that has decayed is 7/8

### Problem 5

The half-life of a radioactive element is 24 hours. Calculate the fraction of the original element that would have disintegrated in 96 hours.

The final answer to the above question is 15/16

#### Explanation

I will apply two methods to solve the problem

##### Method 1: Conventional Method

After 24 hrs, 1/2 disintegrate, 1/2 remain

Another 24 hrs, 1/4 decay, 1/4 remain

Next 24 hrs, 1/8 decay, 1/8 remain

The next 24 hrs, 1/16 decay, 1/16 remain

Fraction disintegrate = sum of decayed fractions = 1/2 + 1/4 + 1/8 + 1/16 = 15/16

##### Method 2: Zhepwo Method

Half-life, T = 24 hrs

Decay time, t = 96 hrs

Fraction disintegrated, fd = ?

Solution

T = t/n

This implies that

n = t/T = 96/24 = 4

R = 2n = 24 = 16

fd = (R – 1) / R = (16 – 1) / 16 = 15/16

Therefore, the fraction of the original element that would have disintegrated in 96 hours is 15/16.

### Problem 6

A radioactive isotope has a half-life of 20 hours. What fraction of the original radioactive nuclei will remain after 80 hours?

The final answer to the above question is 1/16

#### Explanation

We will still use the two methods to find the answer to the problem

##### Method 1: Conventional Method

Let n be the original number of nuclei

After 20 hours, n/2 disintegrates and n/2 remains

The next 20 hours, 1/2 of n/2 disintegrates and n/4 remains

Next 20 hrs, 1/2 of n/4 decayed, and n/8 remains

After another 20 hrs, 1/2 of n/8 decayed, and n/16 remains

##### Method 2: Zhepwo Method

Half-life, T = 20 hrs

Decay time, t = 80 hrs

Number of half-lives, n = t/T = 80/20 = 4

Disintegration ratio, R = 2n = 24 = 16

Fraction remaining, fr = 1/R = 1/16

We can also use an alternative method

T = 20, t = 80

Fraction remaining, fr = 1 / (2t/T) = 1 / (280/20) = 1 / 24 = 1 / 16

Therefore, after 80 hours. The fraction of the original number remaining would be 1/16

### Problem 7

Two radioactive elements A and B have half-lives of 100 and 50 years respectively. Samples of A and B initially contain equal number of atoms. What is the ratio of remaining atoms of A to that of B after 200 years?

The ratio of remaining atoms of A to that of B after 200 years is 4:1

#### Explanation

You can employ any of the two methods below to arrive at your answer:

##### 1. Conventional Method:

When we allow n to be as the original number of the nuclei

Sample A: Half-life = 100 years

After 100 years, n/2 disintegrates and n/2 remains

Additionally, after 100 years, n/4 disintegrates and n/4 remains

Thus, after 200 years, the fraction of the original number that will remain is 1/4

Sample B: Half-life = 50 years

After 50 years, n/2 disintegrate, n/2 remain

Another 50 years, n/4 disintegrate and n/4 will remain

50 years more, n/8 will decay, n/8 remain

After 50 years, n/16 decay, n/16 remain

Therefore, after 200 years, fraction of atoms that will remain is 1/16

Ratio of A:B = 1/4 : 1/16

We will now multiply both sides by 16 to obtain:

The ratio of A:B = 4 : 1

##### 2. Zhepwo Method

T is the Half-life for A and B which is 100 and 50 respectively

And t is the decay time for A and B which is 200 and 200 respectively

The number of half-lives n = t/T which is A = 200/100 = 2, and B = 200/50 = 4

Disintegrating ratio R = 2n and it will give us 22 = 4 and 24 = 16

The initial number of atoms, N1 = 1

Thus, we have A = 16 (1/4) = 4, and B = 16 (1/16) = 1

Hence we have a ratio of A:B = 4:1

### Problem 8

A radioactive substance has a half-life of 80 days. If the initial number of atoms in the sample is 6 x 1010, how many atoms would remain at the end of 320 days?

The number of atoms that would remain at the end of 320 days is 3.8 x 109 atoms

#### Explanation

T = 80; t = 320; and N1 = 6 x 1010

We will also use the formula N2 = (N1 / 2t/T)

Therefore, by substituting our data into the above formula, we will have:

N2 = (N1 / 2t/T) = (6 x 1010) / 2320/80 = 6 x 1010 / 24 = 6 x 1010 / 16 = 3.8 x 109 atoms

### Problem 9

A percentage of the original nuclei of a sample of a radioactive substance left after 5 half-lives is?

The percentage left is 3%

#### Explanation

Number of hlaf-lives (n) = 5

original amount or fraction (N1) = 1

Disintegrating ratio, R = 2n = 25 = 32

Fraction remaining, fr = 1/R

The formula we will apply is:

Percentage of the original left = [amount left (fraction remaining) / original amount] x 100

Therefore,

Percentage of the original left = (fr / N1) x 100 = [(1/32) / 1] x 100 = [1/32] x 100 = 3.125 = 3%

### Problem 10

A radioactive substance of mass 768 grams has a half-life of 3 years. After how many years does this substance leave only 6 grams undecayed?

#### Explanation

Half-life T = 3 years

Initial mass present, N1 = 768 g

Final mass remaining, N2 = 6 g

Decay time, t = ?

The disintegration ratio, R = N1 / N2 = 768 / 6 = 128

from the formula below:

T = t / (log2R)

Making t subject of the formula and substituting our values, we will have:

t = T x log2R = 3 x log2128 = 3 x 7 = 21 years

### Problem 11

An element whose half-life is 10 days is of mass 12 grams. Calculate the time during which 11.25 grams of the element would have decayed.

The final answer to this question is 40 days

#### Explanation

Half-life, T = 10 days

Initial mass present, N1 = 12 g

The mass of the element decayed, Nd = 11.25 grams

Nd = N1 – N2

Thus, N2 = N1 – Nd = 12 – 11.25 = 0.75 grams

R = N1 / N2 = 12 / 0.75 = 16

Applying the formula T = t / (log2R) and making t the subject of the formula, we will have:

t = T x log2R = 10 x log216 = 10 x 4 = 40 years

### Problem 11

A radioactive element decreases in mass from 100g to 15g in 6 days. What is the half-life of the radioactive material?

The half-life of the radioactive element is 2.2 days

#### Explanation

Initial mass present, N1 = 100g

Final mass present, N2 = 15g

Decay time, t = 6 days

Disintegration ratio, R = N1 / N2 = 100 / 5 = 6.67

The half-life formula T = t / (log2R) = 6 / log26.67

Therefore, we can now say that

T = (6 x log2) / log26.67 = (6 x 0.30103) / 0.824 = 2.2 days

### Problem 12

The time it will take a certain radioactive material with a half-life of 50 days to reduce 1/32 of its original number is

The final answer to the above question is 250 days

#### Explanation

Half-life, T = 50 days

Decay time, t = ?

Disintegration ratio, R = 32

Fraction remaining, fr = 1/R = 1/32

Using the formula, we will obtain

t = T x log2R = 50 x log232 = 50 x 5 = 250 days

### Problem 13

A radioactive substance has a half-life of 3 minutes. After 9 minutes, the count rate was observed to be 200, what was the count rate at zero time?

Solution

Note that count rate at zero time is the same as the initial count rate

9 min ===> count rate 200

6 min ===> count rate (200 x 2) = 400

3 min ===> count rate (400 x 2) = 800

0 min ===> count rate (800 x 2) = 1600

Therefore, count rate at zero time is 1600

### Problem 14

The count rate of a radioactive material is 800 count/min. If the half-life of the material is 4 days, what would the the count rate be 16 days later?

Solution

Initial count rate = 800 count/min

Half-life = 4 days

After 4 days ==> (1/2) x 800 = 400 count/min

8 days after ==> (1/2) x 400 = 200 count/min

12 days after ==> (1/2) x 200 = 100 count/min

4 days after ==> (1/2) x 100 = 50 count/min

Therefore, the count rate 16 days later is 50 count/min

### Problem 15

The half-life of a radioactive source is 1 minute. If a rate meter connected to the source registers 200μA at a given time, what would be its reading after 3 minutes?

A rate meter measures the count rate of radioactive of radioactive substance

Halff-life = 1 minute

1 minutes after, rate meter reads (1/2) x 200μ = 100μ

2 minutes after, rate meter reads (1/2) x 100μ = 50μ

3 minutes after, rate meter reads (1/2) x 50μ = 25μ

Therefore, rate meter reading after 3min is 25μ.

### Problem 16

In 90 seconds, the mass of a radioactive element reduces to 1/32 of its original values. Determine the half-life of the element.

Solution

Decay time, t = 90 seconds

Fraction of initial mass remaining, Fr = 1/32

Therefore, by equating the above formulas, we will have

1/R = 1/32

This implies that R = 32

Hence, Half-life (T) = t / (log2R) = 90 / (log232) = 90 / 5 = 18 seconds

### Problem 17

A radioactive substance has a half-life of 3 days. If a mass of 1.55 g of this substance is left after decaying for 15 days, determine the original value of the mass.

Solution

Half-life (T) = 3 days2

Decay time, t = 15 days

Final mass remaining, N2 = 1.55 g

T = t/n

Therefore, n = t/T = 15/3 = 5

Disintegrating ratio, R = 2n = 25 = 32

R = N1 / N2

Hence, N1 = RN2 = 32 x 1.55

Thus, the original mass N1 =49.6 grams

### Problem 18

In 90 seconds, the mass of a radioactive element reduces to 1/16 of its original value. Determine the half-life of the element.

Solution

Decay time t = 90 seconds, Fraction of mass remaining, Fr = 1/16

But Fr = 1/R

Therefore, 1/R = 1/16, and the value of R = 16

Half-life (T) = t / (log2R) = t / (log216) = 90/4 = 22.5 seconds