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.

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?

Correct Answer- (b.) 0.125 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?

Correct Answer- (a.) 75 minutes

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?

Correct Answer– (c.) 132 seconds

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

Correct Answer- (a.) 8.1 hours

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

Correct Answer- (b.) 164 seconds

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?

Answer.

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

\(\beginn=\frac>\end \)

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) =?

\(\beginN = \frac>>\end \)

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 10 23 atoms at the start, how many atoms would be present after 20.0 days?

Answer.

Half-life = 3.6 days

Initial atoms = 6.02 ×10 23 atoms

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

\(\beginN = N_\times \frac\times \frac>\end \) \(\beginN = 6.02\times 10^\times \frac\times \frac=1.28\times 10^\end \)

Thus, the number of atoms available is 1.28 × 10 22 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.

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

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?

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.

Answer. The amount that remains

n log 0.5 = log 0.53125

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?

Answer. (1/2) n = 0.01

n log 0.5 = log 0.01

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?

Answer.

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?

Answer.

The afraction amount remaining will be-

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.

Answer.

If you lose 75%, then 25% remains.

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 14 C is 5730 years. An archaeological artifact containing wood had only 80% of the 14 C 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

\(\begint=\fraclog\frac<\left [ R \right ]_<0>><\left [ R \right ]>\end \) \(\begint=\frac>log\frac\end \)

= 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?

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

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?

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