Question

1 g radius is reduced by 2.1 mg in 5 years by alpha decay, calculate the half-life period.

Answer 1

Hint: We can solve the above problem by using the radioactive decay formula as given below:
$N={{N}_{o}}{{e}^{-\lambda t}}$
Where ${{N}_{o}}$ is the initial quantity of substance.
$N$ quantity remained after time t.
$\lambda $ is decay constant.

Complete step-by-step answer:

Given data is
The initial weight of radium is = 1g
Weight that is reduced is = 2.1mg
Therefore the final weight of the radius is initial weight reduced weight is \[1g-2.1mg\]
\[\Rightarrow 1g-2.1\times \mathop{10}^{-3}g\] $\left\{ \because 1mg={{10}^{-3}}g \right\}$
\[\Rightarrow 0.9979g\approx 0.998g\]
Therefore the final weight of radium is 0.998 g.

Since, we have
Initial weight $N_0$ = 1g
Final weight N(t) = 0.998 g
The time required to reduce by 2.1 mg is 5 years.
We can find out \[\text{ }\!\!\lambda\!\!\text{ }\] (decay Constant) by using radioactive decay formula.
\[N=\mathop{N}_{0}\mathop{e}^{\text{- }\!\!\lambda\!\!\text{ t}}\]
We can rearrange it to find the value of \[\text{ }\!\!\lambda\!\!\text{ }\]
\[\Rightarrow \left( \dfrac{N}{\mathop{N}_{0}} \right)=\mathop{e}^{\text{- }\!\!\lambda\!\!\text{ t}}\]
\[\Rightarrow \text{ln}\left( \dfrac{N}{\mathop{N}_{0}} \right)=-\text{ }\!\!\lambda\!\!\text{ t}\]
\[\Rightarrow -\text{ln}\left( \dfrac{N}{\mathop{N}_{0}} \right)=\text{ }\!\!\lambda\!\!\text{ t}\]
\[\Rightarrow \text{ln}\left( \dfrac{\mathop{N}_{0}}{N} \right)=\text{ }\!\!\lambda\!\!\text{ t}\] $\left\{ \because -\ln \left( \dfrac{a}{b} \right)=\ln \left( \dfrac{b}{a} \right) \right\}$
Decay constant \[\text{ }\!\!\lambda\!\!\text{ }=\dfrac{1}{t}\ln \left( \dfrac{\mathop{N}_{0}}{N} \right)\]
\[\Rightarrow \text{ }\!\!\lambda\!\!\text{ }=\dfrac{1}{5}\ln \left( \dfrac{1}{0.998} \right)\]

Since we have for half-life there is a relation between decay constant and half life period as given below:
$\Rightarrow {{t}_{{}^{1}/{}_{2}}}=\dfrac{0.693}{\lambda }$
Where ${{t}_{{}^{1}/{}_{2}}}$ is half life period of substance.
On substituting value of decay constant \[\text{ }\!\!\lambda\!\!\text{ }=\text{4}\text{.004005}\times \text{1}{{\text{0}}^{-4}}years\].
$\Rightarrow {{t}_{{}^{1}/{}_{2}}}=\dfrac{0.693}{4.004005\times {{10}^{-4}}year{{s}^{-1}}}$
$\Rightarrow {{t}_{{}^{1}/{}_{2}}}=1730.767years$
So the time required for the 1 g radium which reduces by 2.1 mg in years will take 1730.767 years for decay to half of its weight.

Note: As we used the relation between the decay constant and half life period which we got from radioactive decay formula.
Let’s consider radioactive decay formula
$N={{N}_{o}}{{e}^{-\lambda t}}$
When $t={{t}_{{}^{1}/{}_{2}}}$ , $N=\dfrac{{{N}_{0}}}{2}$ because for half life period remaining amount of substance is half of initial amount of substance.
$\Rightarrow \dfrac{{{N}_{0}}}{2}={{N}_{0}}{{e}^{-\lambda {{t}_{{}^{1}/{}_{2}}}}}$
$\Rightarrow \dfrac{1}{2}={{e}^{-\lambda {{t}_{{}^{1}/{}_{2}}}}}$
We can write it in terms of natural logarithmic function as
$\Rightarrow \ln \left( \dfrac{1}{2} \right)=-\lambda {{t}_{{}^{1}/{}_{2}}}$
$\Rightarrow -\ln \left( \dfrac{1}{2} \right)=\lambda {{t}_{{}^{1}/{}_{2}}}$
$\Rightarrow \ln \left( 2 \right)=\lambda {{t}_{{}^{1}/{}_{2}}}$ $\left\{ \because -\ln \left( \dfrac{a}{b} \right)=\ln \left( \dfrac{b}{a} \right) \right\}$
$\Rightarrow {{t}_{{}^{1}/{}_{2}}}=\dfrac{\ln (2)}{\lambda }$
$\Rightarrow {{t}_{{}^{1}/{}_{2}}}=\dfrac{0.693}{\lambda }$