5- Calculate the fraction of 14-C remaining after a decay interval of three half

Question

5- Calculate the fraction of 14-C remaining after a decay interval of three half-lives? 6- calculate the decay interval when: a) 91% of the 14-C remains b) 9% of the 14-C remains 4. "Chas a half life of 5730 years. Calculate the fraction (or percen) of "c remaining after an interval of 2000 years. 5. Calculate the fraction (or percent) of "eC remaining after a decay interval of three half-lives. 6. Calculate the decay interval ("age) when: a) 91% of the "C remains b) 9% of the "C remains 7. Calculate the age of a chemical syster containing a radioactive nuclide P whose halflife is 5.00 x 10 years, given that the ratio D /P equals

Explanation / Answer

4. Let N0 be the initial amount of 14C6 and let N be the amount left after time t.

As per the radioactive decay law,

log (N/N0) = -0.693*t/t1/2 where t = time and t1/2 = half life.

Plug in t = 2000 years and t/12 = 5730 years and obtain

log (N/N0) = -0.693*(2000 years)/(5730 years) = -0.24188

===> N/N0 = 10^(-0.24188) = 0.5729 0.57

The fraction of 14C6 remaining after 2000 years = 0.57 (ans).

5. Half-life is defined as the time required by a radioactive element to decay to one half of its initial concentration.

For 14C, let we have N as the initial amount of the element.

After 1 half-life, amount decayed = N/2; amount remaining = (N – N/2) = N/2

After 2 half-lives, amount decayed = ½*N/2 = N/4; amount remaining = (N/2 – N/4) = N/4

After 3 half-lives, amount decayed = ½*N/4 = N/8; amount remaining = (N/4 – N/8) = N/8

Therefore, fraction remaining after three half-lives = (N/8)/(N) = 1/8 = 0.125 (ans).

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