
Activity of radioactive elements decreased to one third of original activity \[{R_0}\] in 9 years. After further 9 years, its activity will
A. \[{R_0} \\ \]
B. \[\dfrac{2}{3}{R_0} \\ \]
C. \[\dfrac{{{R_0}}}{9} \\ \]
D. \[\dfrac{{{R_0}}}{6}\]
Answer
164.7k+ views
Hint:The activity or decay rate is defined as the number of nuclei decayed per second and its SI unit is Becquerel (Bq). The activity is a positive quantity and it is similar to that of the law of radioactive decay. It also expresses exponential decay characteristics.
Formula used:
The activity of the given radioactive element is,
\[R = {R_0}{e^{ - \lambda t}}\]
Where, \[R\] - activity of the radioactive sample
\[{R_0}\] - initial activity,
\[\lambda \] - decay or disintegration constant
\[t\] - time period
Complete step by step solution:
The activity or the decay rate is the number of radioactive substances disintegrated per unit time. The activity also shows exponential decay behaviour as the law of radioactive decay. The activity of the radioactive element is given by,
\[R = {R_0}{e^{ - \lambda t}}\]
The SI unit of activity (R) is Becquerel (Bq) and it is equal to one disintegration per second. The activity also has another commonly used unit called Curie (Ci).
\[1\,Ci = 3.7 \times {\rm{1}}{{\rm{0}}^{10}}Bq\]
Given, \[{R_0}\] = initial activity
Activity after 9 years = \[\dfrac{{{R_0}}}{3}\]
To find, activity after another 9 years ,\[R' = ?\]
The activity of the given sample is,
\[R = {R_0}{e^{ - \lambda t}}\]
During the first 9 years (t = 9 years),
\[\dfrac{{{R_0}}}{3} = {R_0}{e^{ - 9\lambda }} \\ \]
\[\Rightarrow \dfrac{1}{3} = {e^{ - 9\lambda }} \\ \]
During the next 9 years (t = 9 years),
\[R' = {{\mathop{\rm Re}\nolimits} ^{ - 9\lambda }} \\ \]
\[\Rightarrow R' = \dfrac{{{R_0}}}{3}{e^{ - 9\lambda }} \\ \]
\[\Rightarrow R' = \dfrac{{{R_0}}}{3} \times \dfrac{1}{3} \\ \]
\[\therefore R' = \dfrac{{{R_0}}}{9}\]
Hence, the correct answer is option C.
Note : Here, when the activity during the second 9 years is calculated ( \[R' = {{\mathop{\rm Re}\nolimits} ^{ - 9\lambda }}\]), R is being used instead of \[{R_0}\] because in the period of second 9 years, the initial activity will be the activity which is left by the first 9 years, hence \[{R_0}\] is replaced by R.
Formula used:
The activity of the given radioactive element is,
\[R = {R_0}{e^{ - \lambda t}}\]
Where, \[R\] - activity of the radioactive sample
\[{R_0}\] - initial activity,
\[\lambda \] - decay or disintegration constant
\[t\] - time period
Complete step by step solution:
The activity or the decay rate is the number of radioactive substances disintegrated per unit time. The activity also shows exponential decay behaviour as the law of radioactive decay. The activity of the radioactive element is given by,
\[R = {R_0}{e^{ - \lambda t}}\]
The SI unit of activity (R) is Becquerel (Bq) and it is equal to one disintegration per second. The activity also has another commonly used unit called Curie (Ci).
\[1\,Ci = 3.7 \times {\rm{1}}{{\rm{0}}^{10}}Bq\]
Given, \[{R_0}\] = initial activity
Activity after 9 years = \[\dfrac{{{R_0}}}{3}\]
To find, activity after another 9 years ,\[R' = ?\]
The activity of the given sample is,
\[R = {R_0}{e^{ - \lambda t}}\]
During the first 9 years (t = 9 years),
\[\dfrac{{{R_0}}}{3} = {R_0}{e^{ - 9\lambda }} \\ \]
\[\Rightarrow \dfrac{1}{3} = {e^{ - 9\lambda }} \\ \]
During the next 9 years (t = 9 years),
\[R' = {{\mathop{\rm Re}\nolimits} ^{ - 9\lambda }} \\ \]
\[\Rightarrow R' = \dfrac{{{R_0}}}{3}{e^{ - 9\lambda }} \\ \]
\[\Rightarrow R' = \dfrac{{{R_0}}}{3} \times \dfrac{1}{3} \\ \]
\[\therefore R' = \dfrac{{{R_0}}}{9}\]
Hence, the correct answer is option C.
Note : Here, when the activity during the second 9 years is calculated ( \[R' = {{\mathop{\rm Re}\nolimits} ^{ - 9\lambda }}\]), R is being used instead of \[{R_0}\] because in the period of second 9 years, the initial activity will be the activity which is left by the first 9 years, hence \[{R_0}\] is replaced by R.
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