The half-life of a radioactive isotope represents the average time it would take

Question

The half-life of a radioactive isotope represents the average time it would take half of a collection of this type of nucleus to decay. For example, you start with a sample of 1000 Oxygen-15 (15O) nuclei, which has a half-life of 122 seconds. After 122 seconds, half of the 15O nuclei will have decayed into Nitrogen-15 (15N) nuclei. After another 122s, half of the remaining Oxygen nuclei will have also decayed, and so on.
A) Suppose you start with 2.10×10315O nuclei and zero 15N nuclei. How many 15O nuclei remain after 122 s has passed?

B) How many 15N nuclei are there after 122 s has passed?

C) How many 15O nuclei remain after 244 s has passed?

D) How many 15N nuclei are there after 244 s has passed?

E) Suppose you start with 6.80×103 Carbon-14(14C) nuclei. 14C has a half-life of 5730 years and decays into Nitrogen-14(14N) via a beta decay. How much time has passed if you are left with 3.40×10314C nuclei? (The units for years is 'yr'.)

F) How much time has passed if you are left with 1.70×10314C nuclei?

Explanation / Answer

A. Initial amount of 15O Nuclie = 2.1 x 103 = 2100

After 122s, i.e. half life, half of 15O nuclie decayed to 15 N nuclie.

So remaining 15O nuclie = 2100/2= 1050

B. After 122s, number of 15N nuclie are 1050

C.At the beginning of 122s, no of 15O nuclie = 1050

After passing one half life, i.e. 244s, no of 15 O nuclie =1050/2 =525

D. After 244s, no of 15 N nuclie = no of 15O nuclie decayed = 1050 +525 = 1575

E. The initial amount of carbon-14 = 6.8 x 103

Final amunt of carbon-14 = 3.4 x 103

Amount of carbon-14 left = 3.4 x 103 / 6.8 x 103 =0.5

i.e. half of the initial amount.

so time passed = 1 half life = 5730 years

F. The initial amount of carbon-14 = 6.8 x 103

Final amunt of carbon-14 = 1.7 x 103

Amount of carbon-14 left = 1.7 x 103 / 6.8 x 103 =0.25

i.e.half of half of the initial amount.

so time passed = 2 half life = 5730 x 2 = 11460 years

all the best in the course work

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