Answer:
0.9715 Fraction of Pu-239 will be remain after 1000 years.
Explanation:
[tex]\lambda =\frac{0.693}{t_{\frac{1}{2}}}[/tex]
[tex]A=A_o\times e^{-\lambda t}[/tex]
Where:
[tex]\lambda [/tex]= decay constant
[tex]A_o[/tex] =concentration left after time t
[tex]t_{\frac{1}{2}}[/tex] = Half life of the sample
Half life of Pu-239 = [tex]t_{\frac{1}{2}}=24,000 y[/tex][
[tex]\lambda =\frac{0.693}{24,000 y}=2.8875\times 10^{-5} y^{-1][/tex]
Let us say amount present of Pu-239 today = [tex]A_o=x[/tex]
A = ?
[tex]A=x\times e^{-2.8875\times 10^{-5} y^{-1]\times 1000 y}[/tex]
[tex]A=0.9715\times x[/tex]
[tex]\frac{A}{A_0}=\frac{A}{x}=0.9715[/tex]
0.9715 Fraction of Pu-239 will be remain after 1000 years.