Answer:
A particular sample contains [tex]3\times 10^{12}atoms [/tex]of chlorine-36. After [tex]9.03 \times 10^5 years[/tex] , there will [tex]3.7533\times 10^{11} [/tex] atoms of chlorine-36 and [tex]2.6246\times 10^{12} [/tex]atoms of argon-36.
Explanation:
Number of isotope chlorine-36 atoms = [tex]N_o=3\times 10^{12} atoms[/tex]
Half-life of chlorine-36 atoms =[tex]t_{\frac{1}{2}}= 3.01\times 10^5 years[/tex]
Number of chlorine-36 atom after time , t = N
t = [tex] 9.03\times 10^5 years[/tex]
[tex]\lambda =\frac{0.693}{3.01\times 10^5 years}[/tex]
[tex]\log[N]=\log[N_o]-\frac{\lambda \times t}{2.303}[/tex]
[tex]\log[N]=\log[3\times 10^{12} atoms]-\frac{0.693\times 9.03\times 10^5 years}{3.01\times 10^5 years\times 2.303}[/tex]
[tex]N=3.7533\times 10^{11} atoms[/tex]
Number of chlorine-36 atom after time = [tex]3.7533\times 10^{11} atoms[/tex]
Number of argon-36 atom after time :
[tex]=N_o-N=3\times 10^{12} atoms=3.7533\times 10^{11} atom[/tex]
[tex]=2.6246\times 10^{12} atoms[/tex]
