Answer:
The fraction of Sr-90 left after 73 years is 0.1726
Explanation:
Initial mass of the isotope = [tex]N_o[/tex]
Time taken by the sample, t = [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
We have:
[tex]t_{\frac{1}{2}}=28.8 years[/tex]
t = (2018 - 1945 )years = 73 years
So, on substituting the values:
[tex]N=x\times e^{-(\frac{0.693}{28.8 years})\times 73 years}[/tex]
Now put all the given values in this formula, we get
[tex]N=N_o\times e^{-1.7566}[/tex]
[tex]N=N_o\times 0.1726[/tex]
[tex]\frac{N}{N_o}=0.1726[/tex]
The fraction of Sr-90 left after 73 years = 0.1726
