Respuesta :
Answer:
11.02 % of an isotope will be left after 45 seconds.
Explanation:
[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 :
Mass of Beryllium-11 radioactive isotope= [tex]N_o=100[/tex]
Mass of Beryllium-11 radioactive isotope after 45 seconds = [tex]N=?[/tex]
t = 45 s
[tex]\lambda[/tex] = rate constant = [tex]4.9 \% s^{-1}=0.049 s^{-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=100\times e^{-0.049 s^{-1}\ties 45 s}[/tex]
[tex]N=11.02 g[/tex]
Percentage of isotope left :
[tex]\frac{N}{N_o}\times 100=\frac{11.02 g}{100 g}\times 100=11.02\%[/tex]
11.02 % of an isotope will be left after 45 seconds.
