
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
162.6k+ 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.
Recently Updated Pages
JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Charging and Discharging of Capacitor

Wheatstone Bridge for JEE Main Physics 2025

Formula for number of images formed by two plane mirrors class 12 physics JEE_Main

In which of the following forms the energy is stored class 12 physics JEE_Main
