Question

A specific radioactive substance follows a continuous exponential decay model. It has a half-life of minutes. At the start of the experiment, is present.

Answer 1

The governing equation for a continuous (exponential) decay is given by,

`y=y_0\cdot e^(kt)`

Here, the amount of substance at the beginning (t=0) is,

`y_0`

(a)

Given that the half-life is 4 minutes,

`\begin{gathered} y=(y_0)/(2) \\ y_0\cdot e^(k(4))=(y_0)/(2) \\ e^(4k)=\frac{1_{}}{2} \\ \ln (e^(4k))=\ln ((1)/(2)) \\ 4k=\ln (2)^(-1) \\ 4k=-\ln (2) \\ k=(-1)/(4)\ln (2) \end{gathered}`

Substitute the value of 'k' in the expression,

`y=y0\cdot e^{((-1)/(4)\ln (2))t}`

It is mentioned that the amount is 118.4 gm at the beginning of the experiment,

`y_0=118.4`

Then the equation becomes,

`y=118.4e^{((-1)/(4)\ln (2))t}`

This is the required formula relating 'y' to 't'.

(b)

The amount corresponding to 15 minutes is calculated as,

`\begin{gathered} y=118.4e^{((-1)/(4)\ln 2)(15)} \\ y\approx118.4(0.074) \\ y\approx8.8 \end{gathered}`

Thus, the amount of substance in 15 minutes is 8.8 gm approximately.