Data analysis

Given data,

Initial amount of substance \((P_{0}) = 22 gms\)

Half-life \((t_{1/2})=8\ days\)

Time period \((t) = 5\ days\)

Amount of substance left after 't' = ?

Solution

The decay of an radioactive substance is exponential.

And the amount of substance (P) left fter 't' days is given by,

\(P=P_{0}e^{-\lambda\ t}\)

Where lambda is decay constant.

Given \(t_{1/2} = 8\ days\)

That is by 8 days, amount left will become half.

\(\Rightarrow\ (P_{0}/2)=P_{0}e^{-\lambda(8)}\)

\(\Rightarrow\ 8\lambda=\ln(2)\)

\(\Rightarrow\ \lambda=(1/8)\ln(2)\)

\(\Rightarrow\ \lambda = 0.0866434\) (rounded to 7 decimals)

Now after 5 days, amount left,

\(\Rightarrow\ P=22(e^{0.0866434(5)})\)

\(= 22 (0.6484198)\)

\(\Rightarrow\ P = 14.2652356\) NAK Hemce amount left after 5 days is 14.265 grams (rounded to 3 decimals)

Given data,

Initial amount of substance \((P_{0}) = 22 gms\)

Half-life \((t_{1/2})=8\ days\)

Time period \((t) = 5\ days\)

Amount of substance left after 't' = ?

Solution

The decay of an radioactive substance is exponential.

And the amount of substance (P) left fter 't' days is given by,

\(P=P_{0}e^{-\lambda\ t}\)

Where lambda is decay constant.

Given \(t_{1/2} = 8\ days\)

That is by 8 days, amount left will become half.

\(\Rightarrow\ (P_{0}/2)=P_{0}e^{-\lambda(8)}\)

\(\Rightarrow\ 8\lambda=\ln(2)\)

\(\Rightarrow\ \lambda=(1/8)\ln(2)\)

\(\Rightarrow\ \lambda = 0.0866434\) (rounded to 7 decimals)

Now after 5 days, amount left,

\(\Rightarrow\ P=22(e^{0.0866434(5)})\)

\(= 22 (0.6484198)\)

\(\Rightarrow\ P = 14.2652356\) NAK Hemce amount left after 5 days is 14.265 grams (rounded to 3 decimals)