It takes approximately 55 mins for the sample to decay to one - fourth of its original amount.
In the given statement is:
λ = 2.5 x 10^-2 [tex]min^{-1}[/tex]is the rate constant of the nucleus
To solve this problem, we use
N(t) = [tex]N_{0}[/tex]e^-λt
and since we are looking for the time it takes to decay the nucleus to one -fourth of its original amount, we set
N(t) = 1/4 [tex]N_{0}[/tex] = 0.25[tex]N_{0}[/tex]
We can substitute this in the decay equation:
0.25[tex]N_{0}[/tex] =[tex]N_{0}[/tex] e^-λt
Cancel the like terms
0.25 = e ^-λt
Take the natural logarithm of both sides:
㏑(0.25) =㏑(e^-λt)
㏑(0.25)= -λt
Solve for the time t:
t = - ㏑(0.25)/λ
Substitute the given rate constant :
t = ㏑(0.25) / 2.5 x 10^-2 [tex]min^{-1}[/tex]
t = -1.3863 /2.5 x 10^-2 [tex]min^{-2}[/tex]
t = 55.452mins
It takes approximately 55 mins for the sample to decay to one - fourth of its original amount.
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