# 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. \[2\sqrt 2 \]

Last updated date: 28th Jan 2023

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Answer

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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.

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