Cobalt-60 is a radioactive element with a half-life of 5.25 years. What fraction of the original sample will be left after 26 years? Ans. (1/32)



Data Given:


Half-life of Cobalt-60 = T₁/₂ = 5.25 years 

Time =26 years


To Find:


Fraction of original radioactive isotope remaining after 26 years = `\frac{N}{Nâ‚’}` = ?


Solution:

Time taken in 1st Half-life of Cobalt-60 = 1 T₁/₂ = 5.25 years

Time taken in 2nd Half-lives of Cobalt-60 = 2 T₁/₂ = 2 ｘ5.25 years  = 10.50 years

Time taken in 3rd Half-lives of Cobalt-60 = 3 T₁/₂ = 3 ｘ5.25 years  = 15.75 years

Time taken in 4th Half-lives of Cobalt-60 = 4 T₁/₂ = 4 ｘ5.25 years  = 21 years

Time taken in 5th Half-lives of Cobalt-60 = 5 T₁/₂ = 5 ｘ5.25 years  = 26.25 years ≈ 26 years

Thus, During 26 years, Five Half-lives (5 T₁/₂ ) are elapsed. 

If Nâ‚’ is the original fraction and the remaining fraction (N) of the nuclide of Nitrogen, then after Five half-lives (5 T₁/₂ ),  the fraction can be found by using the formula

Remaining fraction = Original Fraction ï½˜ `\frac{1}{2^t}`

N = Nâ‚’ ï½˜ `\frac{1}{2^t}`

or

`\frac{N}{Nâ‚’}` =  `\frac{1}{2^t}`

by putting values 

`\frac{N}{Nâ‚’}` =  `\frac{1}{2^5}`

`\frac{N}{Nâ‚’}` =  `\frac{1}{32^{th}}`   ---- Ans

Thus the fraction of the original nuclide of Nitrogen will be`\frac{1}{32^{th}}` after 5 half-lives