The half-life of a radioactive substance is $30$ min. The time (in minutes) taken between $40\% $ decay and $85\% $ decay of the same radioactive substance is:
A. $15$
B. $60$
C. $45$
D. $30$
Answer
598.8k+ views
Hint: Half-life of a radioactive substance is the time taken for its given amount to become half. The ratio of remaining radioactive substance to the initial amount of substance is given as: \[\dfrac{N}{{{N_0}}} = {\left( {\dfrac{1}{2}} \right)^{\dfrac{t}{T}}}\] here, \[N\] is the remaining nuclei and \[{N_0}\] is the initial amount of the nuclei, \[t\] is the time required and \[T\] is the half-time. Using this equation, we can find the time taken between $40\% $ decay and $85\% $ decay of the radioactive substance.
Complete step by step answer:
The ratio of the remaining nuclei to the initial nuclei of a radioactive substance is given as:
\[\dfrac{N}{{{N_0}}} = {\left( {\dfrac{1}{2}} \right)^{\dfrac{t}{T}}}\] --equation \[1\]
As per first given condition, the substance has decayed $40\% $ , therefore \[60\% \] of the substance is remaining:
Let the time taken for this decay be \[{t_1}\] , therefore substituting these values in equation \[1\] we have:
\[\dfrac{N}{{{N_0}}} = \dfrac{{60}}{{100}} = {\left( {\dfrac{1}{2}} \right)^{\dfrac{{{t_1}}}{T}}}\] --equation \[2\]
As per second given condition, the substance has decayed $85\% $ , therefore \[15\% \] of the substance is left. Let the time taken for $85\% $ decay be \[{t_2}\] , substituting these values in equation \[1\] we have:
\[\dfrac{N}{{{N_0}}} = \dfrac{{15}}{{100}} = {\left( {\dfrac{1}{2}} \right)^{\dfrac{{{t_2}}}{T}}}\] --equation \[3\]
Dividing equation equation \[3\] by equation \[2\] , we get
\[\dfrac{{\dfrac{{15}}{{100}}}}{{\dfrac{{60}}{{100}}}} = \dfrac{{{{\left( {\dfrac{1}{2}} \right)}^{\dfrac{{{t_2}}}{T}}}}}{{{{\left( {\dfrac{1}{2}} \right)}^{\dfrac{{{t_1}}}{T}}}}}\]
\[ \Rightarrow \dfrac{{15}}{{60}} = {\left( {\dfrac{1}{2}} \right)^{\dfrac{{{t_2} - {t_1}}}{T}}}\]
\[ \Rightarrow \dfrac{1}{4} = {\left( {\dfrac{1}{2}} \right)^{\dfrac{{{t_2} - {t_1}}}{T}}}\]
\[ \Rightarrow {\left( {\dfrac{1}{2}} \right)^{\dfrac{1}{2}}} = {\left( {\dfrac{1}{2}} \right)^{\dfrac{{{t_2} - {t_1}}}{T}}}\]
Comparing the powers, we get:
\[ 2 = \dfrac{{{t_2} - {t_1}}}{T}\]
We need to find the time (in minutes) taken between $40\% $ decay and $85\% $ decay of the same radioactive substance \[{t_2} - {t_1}\] ,
\[{t_2} - {t_1} = 2 \times T\]
We have half-life \[T = 30\,\min \] , substituting this value we get:
\[\therefore{t_2} - {t_1} = 60\,\min \]
This is the time taken between $40\% $ decay and $85\% $ decay of the same radioactive substance.
Therefore, option B is the correct option.
Note:The time taken for a radioactive nucleus to become half is known as the half-life. To calculate the time taken for decay of radioactive nucleus, the ratio of the remaining substance to the initial amount of the substance is taken. The maximum amount of substance decays in the first half-life.
Complete step by step answer:
The ratio of the remaining nuclei to the initial nuclei of a radioactive substance is given as:
\[\dfrac{N}{{{N_0}}} = {\left( {\dfrac{1}{2}} \right)^{\dfrac{t}{T}}}\] --equation \[1\]
As per first given condition, the substance has decayed $40\% $ , therefore \[60\% \] of the substance is remaining:
Let the time taken for this decay be \[{t_1}\] , therefore substituting these values in equation \[1\] we have:
\[\dfrac{N}{{{N_0}}} = \dfrac{{60}}{{100}} = {\left( {\dfrac{1}{2}} \right)^{\dfrac{{{t_1}}}{T}}}\] --equation \[2\]
As per second given condition, the substance has decayed $85\% $ , therefore \[15\% \] of the substance is left. Let the time taken for $85\% $ decay be \[{t_2}\] , substituting these values in equation \[1\] we have:
\[\dfrac{N}{{{N_0}}} = \dfrac{{15}}{{100}} = {\left( {\dfrac{1}{2}} \right)^{\dfrac{{{t_2}}}{T}}}\] --equation \[3\]
Dividing equation equation \[3\] by equation \[2\] , we get
\[\dfrac{{\dfrac{{15}}{{100}}}}{{\dfrac{{60}}{{100}}}} = \dfrac{{{{\left( {\dfrac{1}{2}} \right)}^{\dfrac{{{t_2}}}{T}}}}}{{{{\left( {\dfrac{1}{2}} \right)}^{\dfrac{{{t_1}}}{T}}}}}\]
\[ \Rightarrow \dfrac{{15}}{{60}} = {\left( {\dfrac{1}{2}} \right)^{\dfrac{{{t_2} - {t_1}}}{T}}}\]
\[ \Rightarrow \dfrac{1}{4} = {\left( {\dfrac{1}{2}} \right)^{\dfrac{{{t_2} - {t_1}}}{T}}}\]
\[ \Rightarrow {\left( {\dfrac{1}{2}} \right)^{\dfrac{1}{2}}} = {\left( {\dfrac{1}{2}} \right)^{\dfrac{{{t_2} - {t_1}}}{T}}}\]
Comparing the powers, we get:
\[ 2 = \dfrac{{{t_2} - {t_1}}}{T}\]
We need to find the time (in minutes) taken between $40\% $ decay and $85\% $ decay of the same radioactive substance \[{t_2} - {t_1}\] ,
\[{t_2} - {t_1} = 2 \times T\]
We have half-life \[T = 30\,\min \] , substituting this value we get:
\[\therefore{t_2} - {t_1} = 60\,\min \]
This is the time taken between $40\% $ decay and $85\% $ decay of the same radioactive substance.
Therefore, option B is the correct option.
Note:The time taken for a radioactive nucleus to become half is known as the half-life. To calculate the time taken for decay of radioactive nucleus, the ratio of the remaining substance to the initial amount of the substance is taken. The maximum amount of substance decays in the first half-life.
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