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# In a sample of radioactive material, what fraction of an initial number of active nuclei will remain undisintegrated after half-life of a half-life of the sample? A.$\dfrac{1}{4}$ B. $\dfrac{1}{{2\sqrt 2 }}$ C.$\dfrac{1}{{\sqrt 2 }}$ D. $2\sqrt 2$ Verified
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Hint: Use the expression of decay constant$\left( \lambda \right)$deduced from the relation between half-life and decay constant in the radioactive decay formula.

Formula used:
$N = {N_0}{e^{ - \lambda t}}$ and $t_{\frac {1}{2}}=\dfrac {\ln 2}{\lambda}$
Where,
N = Number of atoms after time t, ${N_0} =$ Number of atoms at time t= 0
$\lambda =$Decay constant and $t_{\frac {1}{2} }$= half life.

Complete step by step solution:
Given here is a sample of radioactive material and we have to calculate the number of nuclei that will remain after half of the half-life $t_{\frac{1}{2}}$ of the sample.

From the information given in question we know that time $t = \dfrac {t_{\frac {1}{2}}}{2}$.

As we know that decay constant $\left( \lambda \right)$ is related with half life by equation
$t_{\frac {1}{2}}=\dfrac {\ln 2}{\lambda}$.

The decay constant can be expressed as
$\lambda=\dfrac {\ln 2}{t_{\frac {1}{2}}}$.

From the law of radioactive disintegration we have,
$N = {N_0}{e^{ - \lambda t}}\,..........(1)$
Where,
N = Number of atoms after time t, ${N_0} =$ Number of atoms at time t= 0

Substituting $t = \dfrac {t_{1/2}}{2}$ and $\lambda=\dfrac {\ln 2}{t_{\frac {1}{2}}}$ in ${e^{ - \lambda t}}$ we get,
$e^{ - \lambda t}=e^{-\left (\dfrac {\ln 2}{t_{1/2}}\right) \left(\dfrac {t_{1/2}}{2}\right)}= e^{-\dfrac {\ln 2}{2}}$
$e^{ - \lambda t}= e^{-\dfrac {\ln 2}{2}}$

Substituting ${e^{ - \lambda t}} = {e^{ - \left( {\dfrac{{\ln 2}}{2}} \right)}}$ in equation (1) we get,
$N = {N_0}{e^{ - \left( {\dfrac{{\ln 2}}{2}} \right)}}\,.........(2)$

Equation (2) can be expressed as,
$N = {N_0}{e^{ - \dfrac{1}{2}\left( {\ln 2} \right)}}\, \Rightarrow {N_0}{e^{\ln {2^{ - \frac{1}{2}}}}}$
$N = {N_0}{e^{\ln {2^{ - \dfrac{1}{2}}}}}$

Using in the above equation it can be expressed as,
$N = {N_0}\,{2^{ - \left( {\dfrac{1}{2}} \right)}}\,$

Further solving the above equation we get,
$N = {N_0}\,{2^{ - \dfrac{1}{2}}} \Rightarrow \,N\, = \dfrac{{{N_0}}}{{{2^{\dfrac{1}{2}}}}}\,\,\,\,\,\,\,\,\,\,\,$
As ${2^{\frac{1}{2}}} = \sqrt 2$

Above equation can be written as,
$\,N\, = \dfrac{{{N_0}}}{{\sqrt 2 }} \Rightarrow N = \dfrac{1}{{\sqrt 2 }}{N_0}\,\,\,\,\,\,\,\,\,\,\,$

So, after half of the half-life of the given sample amount of undisintegrated nuclei will be $\dfrac{1}{{\sqrt 2 }}$of the initial amount of the given sample.

Therefore, option C is the correct option.

Note: Remaining amount of radioactive sample after half of half life will not be equal to the exact half of $\dfrac{{{N_0}}}{2}$ where, ${N_0}$ is the initial amount of radioactive sample.