Answer:
1)50.007 grams of an isotope will left after 10 years.
1)25.007 grams of an isotope will left after 20 years.
3) 23 half-lives will occur in 40 years.
Explanation:
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][N_o]=100 g[/tex]
[tex]t_{1/2}=10 years[/tex]
[tex]\lambda =\frac{0.693}{t_{1/2}}=\frac{0.693}{10 year}=0.0693 year^{-1}[/tex]
1) mass of isotope left after 10 years:
t = 10 years
[tex]N=N_o\times e^{-\lambda t}[/tex]
[tex]N=100g\times e^{-0.0693 year^{-1}\times 10}=50.007 g[/tex]
2) mass of isotope left after 20 years:
t = 20 years
[tex]N=N_o\times e^{-\lambda t}[/tex]
[tex]N=100g\times e^{-0.0693 year^{-1}\times 20}=25.007 g[/tex]
3) Half-lives will occur in 40 years
Mass of isotope left after 40 years:
t = 40 years
[tex]N=N_o\times e^{-\lambda t}[/tex]
[tex]N=100g\times e^{-0.0693 year^{-1}\times 40}=6.2537 g[/tex]
number of half lives = n
[tex]N=\frac{N_o}{2^n}[/tex]
[tex]6.2537 g=\frac{100 g}{2^n}[/tex]
[tex]n\ln 2=\frac{100 g}{6.2537 g}[/tex]
n = 23
23 half-lives will occur in 40 years.
