Half Life Lab Answer Key

One-half-Life or previously known as the Half-Life Catamenia is one of the common terminologies used in Science to depict the radioactive decay of a item sample or element within a certain period of fourth dimension.

Nonetheless, this concept is also widely used to describe various types of decay processes, peculiarly exponential and not-exponential decay. Autonomously from science, the term is used in medical sciences to stand for the biological half-life of certain chemicals in the human trunk 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 unlike substance. The principle was outset discovered in 1907 by Ernest Rutherford. Information technology is usually represented by the symbol Ug or t_1/two.

Information technology can also be referred to equally 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 chosen the half-life period of the reaction.

Half-Life Chemical science 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 chiliad

b.) 0.125 grand

c.) 0.00125 g

d.) 12.five g

Correct Reply- (b.) 0.125 yard

Q2. Selenium-83 has a one-half-life of 25.0 minutes. How many minutes would information technology take for a ten.0 mg sample to decay and simply 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 have a 100.00g sample of As-81, with a half-life of 33 seconds, to decay to half-dozen.25g?

a.) 122 seconds

b.) 101 seconds

c.) 132 seconds

d.) 22 seconds

Correct Answer– (c.) 132 seconds

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

a.) 8.1 hours

b.) half-dozen.1 hours

c.) five hours

d.) 24 hours

Correct Answer- (a.) 8.ane 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 Reply- (b.) 164 seconds

Q6. The half-life of Zn-71 is 2.iv minutes. If one had 100.0 one thousand at the beginning, how many grams would be left later on seven.2 minutes have elapsed?

Answer.

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

Half-life (t½) = two.4 mins

Fourth dimension (t) = 7.2 mins

Number of half-lives

Number of one-half-lives

\(\begin{array}{l}n=\frac{t}{t_{i/two}}\stop{array} \)

n = 7.2/two.4 = 3

Thus, iii half-lives have passed.

Finally, nosotros will calculate the remaining corporeality. This can be obtained by doing the following:

N₀ (original amount) = 100 thou

(north) = number of half-lives

Amount remaining (N) =?

\(\begin{assortment}{l}N = \frac{N_{0}}{North^{due north}}\end{assortment} \)

N = 100 / 2³

N = 100 / 8

N = 12.v thousand

Equally a effect, the corporeality of Zn-71 remaining after 7.2 minutes is 12.5 thou.

Q7. Pd-100 has a half-life of 3.vi days. If ane had vi.02 ten 10²³ atoms at the outset, how many atoms would be present after 20.0 days?

Respond.

Half-life = 3.vi days

Initial atoms = half dozen.02 ×10²³ atoms

Fourth dimension = 20days

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

\(\begin{assortment}{l}Northward = N_{0}\times \frac{1}{2}\times \frac{t}{t_{1/2}}\end{assortment} \)

\(\begin{assortment}{50}Due north = 6.02\times 10^{23}\times \frac{1}{2}\times \frac{20}{3.6}=ane.28\times x^{22}\end{array} \)

Thus, the number of atoms bachelor is one.28 × ten²²atoms.

Q8. Os-182 has a half-life of 21.5 hours. How many grams of a x.0 gram sample would take rust-covered after exactly three half-lives?

Respond. The corporeality of the radioactive substance that will remain after 3- one-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 rust-covered in three one-half-lives = (10 – 1.25) 1000

= eight.75 1000

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

north = log 0.five / 0.015625 n = 6

Adding of the half-life:

24 days divided by six half-lives equals 4.00 days

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

Respond. The amount that remains

17/32 = 0.53125

(1/2)^north = 0.53125

north log 0.five = log 0.53125

due north = 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 accept for a twoscore gram sample of I-131 (half-life = eight.040 days) to decay to 1/100 of its original mass?

Answer. (1/2)ⁿ = 0.01

n log 0.five = log 0.01

due north = 6.64

6.64 x 8.040 days = 53.4 days

Therefore, it will have 53.4 days to decay to ane/100 of its original mass.

Q12. At fourth dimension zero, there are x.0 grams of W-187. If the half-life is 23.ix hours, how much will be present at the end of one day? Two days? 7 days?

Answer.

24.0 hr / 23.9 hr/one-half-life = one.0042 half-lives

One day = ane half-life; (1/2)^one.0042 = 0.4985465 remaining = 4.98 g

Two days = two half-lives; (one/2)^two.0084 = 0.2485486 remaining = 2.48 grand

Vii days = 7 one-half-lives; (1/2)^7.0234 = 0.0076549 remaining = 0.0765 thousand

Q13. 100.0 grams of an isotope with a one-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)^north = 0.05

n log 0.5 = log 0.05

north = 4.32 half-lives

36.0 hours x 4.32 = 155.six hours

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

Reply.

If you lose 75%, and so 25% remains.

(1/2)^{due north} = 0.25

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

12.26 x 2 = 24.52 years

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

Q15. The one-half-life for the radioactivity of ¹⁴C is 5730 years. An archaeological artifact containing woods had only fourscore% of the ^fourteenC institute in a living tree. Guess the age of the sample.

Reply. Decay constant, m = 0.693/t_1/two = 0.693/5730 years = 1/209 × 10^–4/year

\(\begin{array}{l}t=\frac{2.303}{k}log\frac{\left [ R \correct ]_{0}}{\left [ R \right ]}\end{assortment} \)

\(\begin{assortment}{l}t=\frac{two.303}{ane.209\times 10^{-4}}log\frac{100}{80}\finish{assortment} \)

= 1846 years (approx)

Do Questions on Half-Life

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

a.) 1 hour

b.) i-day

c.) 1 calendar week

d.) 1 month

Q2. If the disuse constant of a radioactive nuclide is 6.93 x ten^–3 sec^–ane, its half-life in minutes is:

a.) 100

b.) ane.67

c.) 6.93

d.) 50

Q3. A get-go-order reaction takes 40 min for 30% decomposition. Calculate t_i/2.

Q4. What will be the time for l% completion of a first-guild 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 commencement-guild reaction was completed in 70 min?

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Zero Order Reaction