Answer:
Percentage of the isotope left is 75.87 %.
Explanation:
Initial mass of the isotope = 1 mg
Time taken by the sample, t = [tex]t_{\frac{1}{2}}=123 years[/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]\lambda =\frac{0.693}{123 year}=0.005635 year^{-1}[/tex]
[tex]N=N_o\times e^{-\lambda \times t}[/tex]
Now put all the given values in this formula, we get
[tex]N=1 mg\times e^{-0.005634 year^{-1}\times 49 years}[/tex]
[tex]N=0.7587 mg[/tex]
Percentage of the isotope left:
[tex]\frac{N}{N_o}\times 100[/tex]
=[tex]\frac{0.7587 mg}{1 mg}\times 100[/tex]
Percentage of the isotope left is 75.87 %.
