The fraction of the original mass of the radioactive isotope with a half-life of 10 years that will remain after 50 years is 1/32.
We can find the fraction of the original mass with the exponential decay equation:
[tex] N(t) = N_{0}e^{-\lambda t} [/tex] (1)
Where:
N(t): is the amount of radioactive isotope at time t
N₀: is the initial amount of radioactive isotope
λ: is the decay constant
t: is the time = 50 y
We can find the decay constant as follows:
[tex] \lambda = \frac{ln(2)}{t_{1/2}} [/tex] (2)
Where:
[tex]t_{1/2}[/tex]: is the half-life of the isotope = 10 y
The decay constant is (eq 2):
[tex] \lambda = \frac{ln(2)}{t_{1/2}} = \frac{ln(2)}{10 y} = 0.069 y^{-1} [/tex]
Now, the fraction of the original amount is (eq 1):
[tex] \frac{N(t)}{N_{0}} = e^{-\lambda t} = e^{-0.069 y^{-1}*50 y} = 0.0317 [/tex]
Since we need to calculate the fraction of the original mass, after some algebraic operations we have:
[tex] \frac{N_{0}}{N(t)} = 32 [/tex]
[tex] N(t) = \frac{1}{32}N_{0} [/tex]
Therefore, the fraction of the original mass that will remain is 1/32.
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I hope it helps you!
