Question

Potassium-40 has a half-life of approximately 1.25 billion years. approximately how many years will pass before a sample of potassium-40 contains one-sixteenth the original amount of parent isotope?

Answer 1

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.

Answer 2

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

Step-by-step explanation:

Initial amount of an isotope =
`N_o`

Amount left after the radio decay = N=
`(N_o)/(16)=0.0625 N_o`

Decay constant=
`\lambda`

Half life of the sample:
`t_{(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[(N_o)/(16)]=\ln[N_o]-0.5544 \text{billion years}^(-1)* T`

`\ln[(1)/(16)]=-0.5544 \text{billion years}^(-1)* T`

`-2.7725=-0.5544\text{billion years}^(-1)* 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.