Answer :
The half life for this radioactive substance is 40.76 years
As we know that radioactive decay is a first order reaction
For first order reaction, we can write
N = [tex]N^{0}[/tex][tex]e^{-kt}[/tex]
Where N = Amount of substance at time ‘t’
N0 = Initial amount of substance
K = rate constant for first order reaction
t = time in years
Now, as per the question,
This substance is decaying at a rate of 1.7% that means 98.3% will remain after decaying.
For time t = 1 years , N/[tex]N^{0}[/tex] = 0.983
Therefore, we have
N/[tex]N^{0}[/tex] = [tex]e^{-kt}[/tex]
0.983 = [tex]e^{-k }[/tex]
-k = ln(0.983)
-k = -0.017
K = 0.017 [tex]year^{-1}[/tex]
Now for half life, We can write N/[tex]N^{0}[/tex] = 0.5
Therefore, N/[tex]N^{0}[/tex] = [tex]e^{-kt}[/tex]
0.5 = [tex]e^{-0.017t}[/tex]
ln(0.5) = -0.017t
-0.693 = -0.017t
t = 0.693/0.017 = 40.76 years
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