Question

How do you find the exponential form of half-life period?

Answer 1

Hint:To find the exponential form for the half-life period, we should know the formula to find the
decay constant and then we can able to derive the formula to find the exponential form of half-life
period.

Complete step by step solution:
Half-life period is the time taken by a substance to disintegrate half of its quantity.

The small number of particles $dN$ decayed over the small period of time $dt$ can be written as,
$\dfrac{{dN}}{{dt}} = - \lambda N$

The negative sign indicates that the amount of quantity is decreasing over the particular interval of time and the $\lambda $ be the decay constant.
$dN = - \lambda Ndt$

Integrating the above equation, we get,
$\int {\dfrac{1}{N}dN = - \lambda \int {dt} } $
$\ln N + {C_0} = - \lambda t + {C_1}$

where ${C_0}$ and ${C_1}$ are constant,
$\ln N = - \lambda t + {C_3}$

where ${C_3} = {C_1} - {C_0}$, if we take exponential form to the above equation we get,
$N = {e^{ - \lambda t + {C_3}}}$

Hence,${e^{{C_3}}} = C$ and the equation becomes,
$N(t) = C{e^{ - \lambda t}}$ … (1)

When $t = 0$ we get,
$N(0) = {N_0} = C{e^0}$

We know that${e^0} = 1$,
${N_0} = C$
${N_0}$ indicates that the amount which is present initially and substituting the value of $C$ in (1)

we get,
$N(t) = {N_0}{e^{ - \lambda t}}$ …(2)

Hence, with the above formula we are able to find the exponential form of half life period.

For e.g., The X atom has $6000$ years as half life period, if we consider this in the above equation.

For that if $N(0) = 100\% $ this will be our initial amount and when$N(6000) = 50\% $, it will be reduced to half of its quantity.

 Let us write
$N(t) = 100{e^{ - \lambda t}}$

Substitute $t = 6000$ we get,
$
N(6000) = 50 = 100{e^{ - \lambda 6000}} \\
\dfrac{1}{2} = {e^{ - 6000\lambda }} \\
$
To eliminate exponential we use natural log,
$\ln \left( {\dfrac{1}{2}} \right) = - 6000\lambda $
$\lambda = \dfrac{{\ln \dfrac{1}{2}}}{{ - 6000}}$

The value of $\lambda = 1.21 \times {10^{ - 4}}$.

Note: When we substitute the value of $\lambda $ in (2) we get,
$N(t) = {N_0}{e^{ - 1.21 \times {{10}^{ - 4}}t}}$

With this formula, whatever may be the time, we are able to find how many particles will be left over any hundred to n number of years’.