Answer:
3.07 seconds is the half-life of the isotope.
Explanation:
Initial mass of an isotope = x
Time taken by the sample, t = 8.40 s
Mass of an isotope decayed= 85.0%
Final mass of an isotope left=(100%-85%)of x= 15.0% of x = 0.15x
Half life of an isotope =[tex]t_{\frac{1}{2}} = ?[/tex]
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
[tex]\lambda[/tex] = rate constant
[tex]0.15x=x\times e^{-(\frac{0.693}{t_{1/2}})\times 8.40 s}\\\\N=N_o\times e^{-0.693}[/tex]
Now put all the given values in this formula, we get
[tex]t_{1/2]=3.07 s[/tex]
3.07 seconds is the half-life of the isotope.
