Calculate the half-life period of a radioactive substance if its activity drops to 1/16th of its initial value in 30 years.

Solution

The activity of a radioactive substance at any time 't' is given by,
$A = A_{o} e^{- λt}$ ... (1)
where,

A is the radioactivity material remaining after time 't,
A_o is the initial activity of the material, and
$λ is the decay constant .$

Given that, $\frac{A}{A_{o}} = \frac{1}{16}$

t = 30 years

Now, putting these values in equation (1), we get

$\frac{1}{16} = e^{- 30 λ} \Rightarrow e^{30 λ} = 16 \Rightarrow 30 λ = \ln 16 = 2 \ln 2 \Rightarrow λ = \frac{\ln 2}{15}$

And, half life of a radioactive substance is given by,

$T_{\frac{1}{2}} = \frac{\ln 2}{λ} Putting the value of λ, we get T_{\frac{1}{2}} = \frac{\ln 2}{\ln \frac{2}{15}} = 15 years$ ,
is the required half-period of the radioactive substance.

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Wave Optics

Calculate the half-life period of a radioactive substance if its activity drops to 1/16th of its initial value in 30 years.

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