
A sample of a radioactive substance undergoes $80\%$ decomposition in 345 minutes. Its half-life is ___________ minutes.
(a) \[\dfrac{\text{ln2}}{\text{ln5}}\times 345\]
(b) \[\dfrac{\text{ln5}}{\text{ln2}}\times 345\]
(c) \[\dfrac{\text{ln5}}{\text{ln4}}\times 345\]
(d) \[\dfrac{\text{ln4}}{\text{ln5}}\times 345\]
Answer
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Hint: Half life of the substance is defined as the time which is required to the radioactive substance so that its half of the atoms get disintegrated or transform to the different substance. This principle was discovered by Ernest Rutherford in 1907. The symbol to express half the life of the substance is Ug.
Complete answer:
The radioactive reactions are considered to be the first order of reactions. The first order of reaction is defined as to be proceeded at the rate which depends linearly on only one of the reactant concentrations. The formula for this is the decay constant is following:
\[\lambda t=\ln (\dfrac{{{N}_{o}}}{N})\]
\[\lambda \] is the decay constant and t is the time and \[{{N}_{o}}\] is the initial number of atoms and N is the left number of atoms
t= 345 minutes
\[{{N}_{o}}\]= 100
$N= 100-80=20$
Substituting the values in formula.
\[\lambda =\dfrac{1}{345}\ln (\dfrac{100}{20})\]
\[\lambda =\dfrac{\ln 5}{345}\]
Now the formula for half life is the following:
\[\lambda =\dfrac{\ln 5}{345}\]
Substituting the value in the formula. We get
\[{{t}_{\dfrac{1}{2}}}=\dfrac{\ln 2}{\dfrac{\ln 5}{345}}=\dfrac{\ln 2\times 345}{\ln 5}=\dfrac{\ln 2}{\ln 5}\times 345\]
So the correct answer for this question is option, ‘(a) \[\dfrac{\text{ln2}}{\text{ln5}}\times 345\]’.
Note: The radioactive substances undergo the process of radioactivity because the nucleus in those substances are unstable. The energy is lost due to the radiation which is emitted during this process. The nucleus is kept by the forces acting on it and they are the force of repulsion and the force of attraction. When the atom increases its instability the size of the nucleus increases as the mass of the nucleus increases because of the high concentration. Due to this reason the compounds such as uranium undergo radioactivity.
Complete answer:
The radioactive reactions are considered to be the first order of reactions. The first order of reaction is defined as to be proceeded at the rate which depends linearly on only one of the reactant concentrations. The formula for this is the decay constant is following:
\[\lambda t=\ln (\dfrac{{{N}_{o}}}{N})\]
\[\lambda \] is the decay constant and t is the time and \[{{N}_{o}}\] is the initial number of atoms and N is the left number of atoms
t= 345 minutes
\[{{N}_{o}}\]= 100
$N= 100-80=20$
Substituting the values in formula.
\[\lambda =\dfrac{1}{345}\ln (\dfrac{100}{20})\]
\[\lambda =\dfrac{\ln 5}{345}\]
Now the formula for half life is the following:
\[{{t}_{\dfrac{1}{2}}}=\dfrac{\ln 2}{\dfrac{\ln 5}{345}}=\dfrac{\ln 2\times 345}{\ln 5}=\dfrac{\ln 2}{\ln 5}\times 345\]
\[\lambda =\dfrac{\ln 5}{345}\]
Substituting the value in the formula. We get
\[{{t}_{\dfrac{1}{2}}}=\dfrac{\ln 2}{\dfrac{\ln 5}{345}}=\dfrac{\ln 2\times 345}{\ln 5}=\dfrac{\ln 2}{\ln 5}\times 345\]
So the correct answer for this question is option, ‘(a) \[\dfrac{\text{ln2}}{\text{ln5}}\times 345\]’.
Note: The radioactive substances undergo the process of radioactivity because the nucleus in those substances are unstable. The energy is lost due to the radiation which is emitted during this process. The nucleus is kept by the forces acting on it and they are the force of repulsion and the force of attraction. When the atom increases its instability the size of the nucleus increases as the mass of the nucleus increases because of the high concentration. Due to this reason the compounds such as uranium undergo radioactivity.
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