Answer:
0.03127 fraction of a sample of this nuclide would remain after 12.5 hrs.
Explanation:
Initial mass of the isotope =[tex]N_o[/tex]
Time taken by the sample, t = 12.5 hour
Formula used :
[tex]N=N_o\times e^{-\lambda t}\\\\\lambda =\frac{0.693}{t_{\frac{1}{2}}}[/tex]
where,
[tex]N_o[/tex] = initial mass of isotope
N = mass of the parent isotope left after the time, (t)
[tex]t_{\frac{1}{2}}[/tex] = half life of the isotope = 2.5 hour
[tex]\lambda[/tex] = rate constant
[tex]N=N_o\times e^{-(\frac{0.693}{2.5 hours})\times 12.5 hour}[/tex]
Now put all the given values in this formula, we get
[tex]N=N_o\times e^{-3.465}[/tex]
[tex]\frac{N}{N_o}=0.03127[/tex]
0.03127 fraction of a sample of this nuclide would remain after 12.5 hrs.
