Answer:
we will have 17.8 % of the original value
Explanation:
As we know that by radioactive decay the total number of nuclei present at any instant of time is given as
[tex]N = N_o e^{-\lambda t}[/tex]
here we need to find the fraction of total number of nuclei present
so we will have
[tex]\frac{N}{N_o} = e^{-\lambda t}[/tex]
so we have
[tex]\lambda = \frac{ln 2}{8.040}[/tex]
now we have
[tex]\frac{N}{N_o} = e^{-\frac{ln 2}{8.040}(20)}[/tex]
[tex]\frac{N}{N_o} = 0.178[/tex]
so we will have 17.8 % of the original value
