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# Half-Life Questions

Half-Life or previously known as the Half-Life Period is one of the common terminologies used in Science to describe the radioactive decay of a particular sample or element within a certain period of time.

However, this concept is also widely used to describe various types of decay processes, especially exponential and non-exponential decay. Apart from science, the term is used in medical sciences to represent the biological half-life of certain chemicals in the human body or in drugs.

Definition: Half-Life is normally defined as the time a radioactive substance (or one half the atoms) needs to disintegrate or transform into a different substance. The principle was first discovered in 1907 by Ernest Rutherford. It is usually represented by the symbol Ug or t1/2.

It can also be referred to as the time taken for half of the reactions to complete or the time at which the concentration of the reactant is reduced to half of its original value is called the half-life period of the reaction.

## Half-Life Chemistry Questions with Solutions

Q1. An isotope of caesium (Cs-137) has a half-life of 30 years. If 1.0g of Cs-137 disintegrates over a period of 90 years, how many grams of Cs-137 would remain?

a.) 1.25 g

b.) 0.125 g

c.) 0.00125 g

d.) 12.5 g

Q2. Selenium-83 has a half-life of 25.0 minutes. How many minutes would it take for a 10.0 mg sample to decay and only have 1.25 mg of it remain?

a.) 75 minutes

b.) 75 days

c.) 75 seconds

d.) 75 hours

Q3. How long does it take a 100.00g sample of As-81, with a half-life of 33 seconds, to decay to 6.25g?

a.) 122 seconds

b.) 101 seconds

c.) 132 seconds

d.) 22 seconds

Q4. What is the half-life of a radioactive isotope if a 500.0g sample decays to 62.5g in 24.3 hours?

a.) 8.1 hours

b.) 6.1 hours

c.) 5 hours

d.) 24 hours

Q5. What is the half-life of Polonium-214 if, after 820 seconds, a 1.0g sample decays to 0.03125g?

a.) 164 minutes

b.) 164 seconds

c.) 64 seconds

d.) 160 minutes

Q6. The half-life of Zn-71 is 2.4 minutes. If one had 100.0 g at the beginning, how many grams would be left after 7.2 minutes have elapsed?

To begin, we’ll count the number of half-lives that have passed. This can be obtained by doing the following:

Half-life (t½) = 2.4 mins

Time (t) = 7.2 mins

Number of half-lives

Number of half-lives

$$\begin{array}{l}n=\frac{t}{t_{1/2}}\end{array}$$

n = 7.2/2.4 = 3

Thus, three half-lives have passed.

Finally, we will calculate the remaining amount. This can be obtained by doing the following:

N0 (original amount) = 100 g

(n) = number of half-lives

Amount remaining (N) =?

$$\begin{array}{l}N = \frac{N_{0}}{N^{n}}\end{array}$$

N = 100 / 23

N = 100 / 8

N = 12.5 g

As a result, the amount of Zn-71 remaining after 7.2 minutes is 12.5 g.

Q7. Pd-100 has a half-life of 3.6 days. If one had 6.02 x 1023 atoms at the start, how many atoms would be present after 20.0 days?

Half-life = 3.6 days

Initial atoms = 6.02 ×1023 atoms

Time = 20days

To calculate the atoms present after 20 days, we use the formula below.

$$\begin{array}{l}N = N_{0}\times \frac{1}{2}\times \frac{t}{t_{1/2}}\end{array}$$
$$\begin{array}{l}N = 6.02\times 10^{23}\times \frac{1}{2}\times \frac{20}{3.6}=1.28\times 10^{22}\end{array}$$

Thus, the number of atoms available is 1.28 × 1022 atoms.

Q8. Os-182 has a half-life of 21.5 hours. How many grams of a 10.0 gram sample would have decayed after exactly three half-lives?

Answer. The amount of the radioactive substance that will remain after 3- half- lives=(½)3 × a,

where a = initial concentration of the radioactive element.

a= 10 g

So, amount of the radioactive substance that remains aftet 3- half-lives=( ½)³x10 = 10/8= 1.25 g.

Therefore, the number of grams of the radioactive substance that decayed in 3 half-lives = (10 – 1.25) g

= 8.75 g

Q9. After 24.0 days, 2.00 milligrams of an original 128.0 milligram sample remain. What is the half-life of the sample?

Answer. The remaining decimal fraction is:

2.00 mg / 128.0 mg = 0.015625

The half-lives that must have expired to get to 0.015625?

(½)n = 0.015625

n log 0.5 = 0.015625

n = log 0.5 / 0.015625 n = 6

Calculation of the half-life:

24 days divided by 6 half-lives equals 4.00 days

Q10. A radioactive isotope decayed to 17/32 of its original mass after 60 minutes. Find the half-life of this radioisotope.

17/32 = 0.53125

(1/2)n = 0.53125

n log 0.5 = log 0.53125

n = 0.91254

Half-lives that have elapsed are therefore, n = 0.9125

60 minutes divided by 0.91254 equals 65.75 minutes.

Therefore, n = 66 minutes

Q11. How long will it take for a 40 gram sample of I-131 (half-life = 8.040 days) to decay to 1/100 of its original mass?

n log 0.5 = log 0.01

n = 6.64

6.64 x 8.040 days = 53.4 days

Therefore, it will take 53.4 days to decay to 1/100 of its original mass.

Q12. At time zero, there are 10.0 grams of W-187. If the half-life is 23.9 hours, how much will be present at the end of one day? Two days? Seven days?

24.0 hr / 23.9 hr/half-life = 1.0042 half-lives

One day = one half-life; (1/2)1.0042 = 0.4985465 remaining = 4.98 g

Two days = two half-lives; (1/2)2.0084 = 0.2485486 remaining = 2.48 g

Seven days = 7 half-lives; (1/2)7.0234 = 0.0076549 remaining = 0.0765 g

Q13. 100.0 grams of an isotope with a half-life of 36.0 hours is present at time zero. How much time will have elapsed when 5.00 grams remains?

The afraction amount remaining will be-

5.00 / 100.0 = 0.05

(1/2)n = 0.05

n log 0.5 = log 0.05

n = 4.32 half-lives

36.0 hours x 4.32 = 155.6 hours

Q14. How much time will be required for a sample of H-3 to lose 75% of its radioactivity? The half-life of tritium is 12.26 years.

If you lose 75%, then 25% remains.

(1/2)n = 0.25

n = 2 (Since, (1/2)2 = 1/4 and 1/4 = 0.25)

12.26 x 2 = 24.52 years

Therefore, 24.52 years of time will be required for a sample of H-3 to lose 75% of its radioactivity

Q15. The half-life for the radioactive decay of 14C is 5730 years. An archaeological artifact containing wood had only 80% of the 14C found in a living tree. Estimate the age of the sample.

Answer. Decay constant, k = 0.693/t1/2 = 0.693/5730 years = 1/209 × 10–4/year

$$\begin{array}{l}t=\frac{2.303}{k}log\frac{\left [ R \right ]_{0}}{\left [ R \right ]}\end{array}$$
$$\begin{array}{l}t=\frac{2.303}{1.209\times 10^{-4}}log\frac{100}{80}\end{array}$$

= 1846 years (approx)

## Practise Questions on Half-Life

Q1. A newly prepared radioactive nuclide has a decay constant λ of 10–6 s–1. What is the approximate half-life of the nuclide?

a.) 1 hour

b.) 1-day

c.) 1 week

d.) 1 month

Q2. If the decay constant of a radioactive nuclide is 6.93 x 10–3 sec–1, its half-life in minutes is:

a.) 100

b.) 1.67

c.) 6.93

d.) 50

Q3. A first-order reaction takes 40 min for 30% decomposition. Calculate t1/2.

Q4. What will be the time for 50% completion of a first-order reaction if it takes 72 min for 75% completion?

Q5. How much time will it take for 90% completion of a reaction if 80% of a first-order reaction was completed in 70 min?

Click the PDF to check the answers for Practice Questions.

## Zero Order Reaction 