Solveeit Logo
Login

Question

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

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

Explanation

Solution

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 dNdN decayed over the small period of time dtdt can be written as,
dNdt=−λN\dfrac{{dN}}{{dt}} = - \lambda N

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

Integrating the above equation, we get,
∫1NdN=−λ∫dt\int {\dfrac{1}{N}dN = - \lambda \int {dt} }
ln⁡N+C0=−λt+C1\ln N + {C_0} = - \lambda t + {C_1}

where C0{C_0} and C1{C_1} are constant,
ln⁡N=−λt+C3\ln N = - \lambda t + {C_3}

where C3=C1−C0{C_3} = {C_1} - {C_0}, if we take exponential form to the above equation we get,
N=e−λt+C3N = {e^{ - \lambda t + {C_3}}}

Hence,eC3=C{e^{{C_3}}} = C and the equation becomes,
N(t)=Ce−λtN(t) = C{e^{ - \lambda t}} … (1)

When t=0t = 0 we get,
N(0)=N0=Ce0N(0) = {N_0} = C{e^0}

We know thate0=1{e^0} = 1,
N0=C{N_0} = C
N0{N_0} indicates that the amount which is present initially and substituting the value of CC in (1)

we get,
N(t)=N0e−λtN(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 60006000 years as half life period, if we consider this in the above equation.

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

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

Substitute t=6000t = 6000 we get,
N(6000)=50=100e−λ6000 12=e−6000λ  N(6000) = 50 = 100{e^{ - \lambda 6000}} \\\ \dfrac{1}{2} = {e^{ - 6000\lambda }} \\\
To eliminate exponential we use natural log,
ln⁡(12)=−6000λ\ln \left( {\dfrac{1}{2}} \right) = - 6000\lambda
λ=ln⁡12−6000\lambda = \dfrac{{\ln \dfrac{1}{2}}}{{ - 6000}}

The value of λ=1.21×10−4\lambda = 1.21 \times {10^{ - 4}}.

Note: When we substitute the value of λ\lambda in (2) we get,
N(t)=N0e−1.21×10−4tN(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’.