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Morgarella [4.7K]
3 years ago
9

Potassium-40 has a half-life of approximately 1.25 billion years. approximately how many years will pass before a sample of pota

ssium-40 contains one-sixteenth the original amount of parent isotope?
Chemistry
2 answers:
mojhsa [17]3 years ago
4 0
In a first-order chemical reaction, the half-life of the reactant is computed by ln(2)/λ/.Where:λ is the reaction rate constant.
In terms of radioactive decay, the half-life is the extent of time after which there is a 50% probability that an atom will have completed a nuclear decay.
Since (1/2)^4 = 1/16, that would require 4 half-lives, which is 5 billion years.
astraxan [27]3 years ago
4 0

Answer: It will take 5 billion years for the parent isotope to get reduced to one-sixteenth of its original amount.

Explanation:

Initial amount of an isotope =N_o

Amount left after the radio decay = N=\frac{N_o}{16}=0.0625 N_o

Decay constant= \lambda

Half life of the sample: t_{\frac{1}{2}}=1.25 billion years

Years pass before a sample of potassium-40 contains one-sixteenth the original amount of parent isotope be ,t= T

\lambda =\frac{0.693}{1.25 \text{billion years}}=0.5544 \text{billion years}^{-1}

N=N_o\time e^{-\lambda t}

\ln[\frac{N_o}{16}]=\ln[N_o]-0.5544 \text{billion years}^{-1}\times T

\ln[\frac{1}{16}]=-0.5544 \text{billion years}^{-1}\times T

-2.7725=-0.5544\text{billion years}^{-1}\times t

T=\frac{-2.7725}{-0.5544\text{billion years}^{-1}}=5.00 \text{billion years}

It will take 5 billion years for the parent isotope to get reduced to one-sixteenth of its original amount.

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