The half-life of ₇¹⁶N is 7.3 s. A sample of this nuclide of nitrogen is observed for 29.2s. Calculate the fraction of the original radioactive isotope remaining after this time.  Ans. (1/16)



Data Given:


Half-life of ₇¹⁶N = T₁/₂ = 7.3 s 

Time =29.2 s


To Find:


Fraction of original radioactive isotope remaining after 29.2 s = `\frac{N}{Nₒ}` = ?


Solution:

Time taken in 1st Half-life of ₇¹⁶N = 1 T₁/₂ = 7.3 s

Time taken in 2nd Half-lives of ₇¹⁶N = 2 T₁/₂ = 2 ｘ7.3 s  =14.6 s

Time taken in 3rd Half-lives of ₇¹⁶N = 3 T₁/₂= 3 ｘ7.3 s  =21.9 s

Time taken in 4th Half-lives of ₇¹⁶N = 4 T₁/₂ = 4 ｘ7.3 s  =29.2 s

Thus, During 29.2 s, Four Half-lives (4 T₁/₂ ) elapsed. 

If Nₒ is the original fraction and the remaining fraction (N) of the nuclide of Nitrogen, then after Four half-lives (4 T₁/₂ ),  the fraction can be found by using the formula

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

N = Nₒ ｘ `\frac{1}{2^t}`

or

`\frac{N}{Nₒ}` =  `\frac{1}{2^t}`

by putting values 

`\frac{N}{Nₒ}` =  `\frac{1}{2^4}`

`\frac{N}{Nₒ}` =  `\frac{1}{16^{th}}`   ---- Ans

Thus the fraction of the original nuclide of Nitrogen will be`\frac{1}{16^{th}}` after 4 half-lives