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 = NNₒ = ?


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 12t

N = Nₒ ｘ 12t

or

NNₒ =  12t

by putting values 

NNₒ =  124

NNₒ =  116th   ---- Ans

Thus the fraction of the original nuclide of Nitrogen will be116th after 4 half-lives