Answer :
The activity due to the Po-218 formed after 20 minutes is 1.62.
The radioactive decay equation for Po-218 is given as;`Po-218 -> ? + alpha
The half-life of Po-218 is given as 3.05 minutes, which means that half of the atoms decay in that time.
We can use this to find the decay constant, λ.`t1/2 = 0.693/λ
`Rearranging to find λ:`
λ = 0.693/t1/2``λ = 0.693/3.05 = 0.2271 min^-1
We can now use this to calculate the activity of Po-218 formed after 20 minutes. The general formula for radioactive decay is;
N = N0 e^(-λt)`Where;N0 = Initial number of radioactive atomsN = Remaining number of radioactive atoms after time tλ = Decay constantt = Time elapsed
We know that N0 = 150, and t = 20 minutes.
Substituting these values and solving for N;
N = N0 e^(-λt)` `= 150 e^(-0.2271 × 20)` `= 150 e^(-4.542)` `= 150 × 0.0108` `= 1.62
Therefore, the activity due to the Po-218 formed after 20 minutes is 1.62.
Learn more about activity with the given link,
https://brainly.com/question/1289354
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