A radioactive isotope of mercury, 197Hg, decays togold, 197Au, with a disintegration constant of 0.0108 h-1. What fraction of a sample will remain at the end of three half-lives (

THE FRACTION OF THE SAMPLE REMAINING AFTER THREE HALF LIVES IS 0.125 OR 125/1000

Explanation:

A radioactive isotope of mercury decay to gold with a disintegration constant of 0.0108 h^-1

To calculate the fraction of sample remaining after three half life, we first calculate the half life of the decay.

Half life = ln 2 / Y

Y = disintegration constant

So therefore,

half life = ln 2 / 0.0108

half life = 0.693 / 0.0108

half life = 64.18 hours.

So a decay occurs after 64.18 hours.

To calculate the fraction remaining after 3 half life:

N(t) = N(o) e ^-Yt

where t = 3 half life

So, N / No = e^-Y ( 3 t1/2)

Since t 1/2 = ln 2 / Y, so we can re-write the formula as:

Nt / No = e^-Y ( 3 ln 2/ Y)

Nt / No = e^-3 ln2

Nt / No = e^-3 * 0.693

Nt / No = e^-2.079

Nt / No = 0.125

So the fraction of the sample remaining after 3 half lives is 125/ 1000 or 0.125

Step-by-step explanation:

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I really really don't know the answer because I didn't learnt that and I don't want to learn this that, thank you for your try, this is what I do to get points, I'm a master mind and I'm smarter than you

I really really don't know the answer because I didn't learnt that and I don't want to learn this that, thank you for your try, this is what I do to get points, I'm a master mind and I'm smarter than you