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. Show 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 SolutionsQ1. 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 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? a.) 75 minutes b.) 75 days c.) 75 seconds d.) 75 hours 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? a.) 122 seconds b.) 101 seconds c.) 132 seconds d.) 22 seconds 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? a.) 8.1 hours b.) 6.1 hours c.) 5 hours d.) 24 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? a.) 164 minutes b.) 164 seconds c.) 64 seconds d.) 160 minutes 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 \(\begin{array}{l}n=\frac{t}{t_{1/2}}\end{array} \) 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? Answer. 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. Answer. The amount that remains 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? Answer. (1/2)n = 0.01 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? 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- 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. Answer. 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-LifeQ1. 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. Recommended VideosZero Order ReactionWhat is the halfhalf-life, in radioactivity, the interval of time required for one-half of the atomic nuclei of a radioactive sample to decay (change spontaneously into other nuclear species by emitting particles and energy), or, equivalently, the time interval required for the number of disintegrations per second of a radioactive ...
How much of a radioactive substance remains after 5 hours if its halfHence, after 5 half lives the quantity that would remain would be: 321 which is 3.
What has a halfThe half-life of phosphorus-32 is 14 days. After 14 days, half of the original amount of phosphorus-32 has decayed, so only half remains.
What is the fraction of atom left after 10 halfA: After 2 half-lives (40 min) 1/22 = 1/4 of the initial activity is left (25%). To have less than one left, one needs a DF < 1/1000. This happens after 10 half-lives because (1/2)10 = 1/1024, so after 200 minutes (3 hrs 20 min).
|