## Half-life Problems And Answers Examples

Here are a few half-life problems and answers examples to help you understand how to solve a question about half-life:

### 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?

#### Answer

**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, N_{1} = 64 g

Final mass, N_{2} = 2 g

We will use the formula

R = N_{1} / N_{2} = 64 / 2 = 32

T = t / (log_{2}R) = 24 / (log_{2}32) = 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?

#### Answer

**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, N_{1} = 1 kg

Mass of the decayed material, N_{d} = ?

**Formula **

T = t/n

This implies that

n = t / T = 24 / 6 = 4

R = 2^{n} = 2^{4} = 16

N_{d} = N_{1} (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?

#### Answer

**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, N_{1} = N

Number of half-lives, n = 3

Also, the number of atoms decayed, N_{d} = ?

**Solution**

R = 2^{n} = 2^{3} = 8

N_{d} = N_{1} (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

#### Answer

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 f_{d} = ?

Disintegration ratio, R = 2^{n} = 2^{3} = 8

f_{d} = 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.

#### Answer

**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, f_{d} = ?

**Solution**

T = t/n

This implies that

n = t/T = 96/24 = 4

R = 2^{n} = 2^{4} = 16

f_{d} = (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?

#### Answer

**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 = 2^{n} = 2^{4} = 16

Fraction remaining, f_{r} = 1/R = 1/16

**We can also use an alternative method **

T = 20, t = 80

Fraction remaining, f_{r} = 1 / (2^{t/T}) = 1 / (2^{80/20}) = 1 / 2^{4} = 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?

#### Answer

**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 = 2^{n} and it will give us 2^{2} = 4 and 2^{4} = 16

The initial number of atoms, N_{1} = 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 10^{10}, how many atoms would remain at the end of 320 days?

#### Answer

**The number of atoms that would remain at the end of 320 days is 3.8 x 10 ^{9} atoms**

#### Explanation

T = 80; t = 320; and N_{1} = 6 x 10^{10}

We will also use the formula N_{2} = (N_{1} / 2^{t/T})

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

**N _{2} = (N_{1} / 2^{t/T}) = (6 x 10^{10}) / 2^{320/80} = 6 x 10^{10} / 2^{4} = 6 x 10^{10} / 16 = 3.8 x 10^{9} atoms**

### Problem 9

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

#### Answer

**The percentage left is 3%**

#### Explanation

Number of hlaf-lives (n) = 5

original amount or fraction (N_{1}) = 1

Disintegrating ratio, R = 2^{n} = 2^{5} = 32

Fraction remaining, f_{r} = 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 = (f _{r} / N_{1}) 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?

#### Answer

**The answer is 21 years**

#### Explanation

Half-life T = 3 years

Initial mass present, N_{1} = 768 g

Final mass remaining, N_{2} = 6 g

Decay time, t = ?

The disintegration ratio, R = N_{1} / N_{2} = 768 / 6 = 128

from the formula below:

T = t / (log_{2}R)

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

**t = T x log _{2}R = 3 x log_{2}128 = 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.

#### Answer

**The final answer to this question is 40 days**

#### Explanation

Half-life, T = 10 days

Initial mass present, N_{1} = 12 g

The mass of the element decayed, N_{d} = 11.25 grams

N_{d} = N_{1} – N_{2}

Thus, N_{2} = N_{1} – N_{d} = 12 – 11.25 = 0.75 grams

R = N_{1} / N_{2} = 12 / 0.75 = 16

Applying the formula T = t / (log_{2}R) and making t the subject of the formula, we will have:

**t = T x log _{2}R = 10 x log_{2}16 = 10 x 4 = 40 years**

*You may also like to read:*