Concepts Of Physics MCQ Edition [Volume 2]
- Exam
- Concepts Of Physics MCQ Edition [Volume 2]
- Subject
- Physics
- Chapter
- The Nucleus
Concepts Of Physics MCQ Edition [Volume 2] · Physics · The Nucleus
In an experiment, a radioactive isotope is generated at a constant...
In an experiment, a radioactive isotope is generated at a constant rate . The isotope has a half-life . Assuming that the generation of the radioactive isotope begins at , determine the number of active nuclei at time .
Options
- A\left(1 + e^{-\lambda t}\right)\lambda = \dfrac{\ln 2}{t_{1/2}})
- B (where )
- C\left(1 - e^{-\lambda t}\right)\lambda = \dfrac{\ln 2}{t_{1/2}})
- D\left(1 - e^{-\lambda t}\right)\lambda = \dfrac{\ln 2}{t_{1/2}})
Correct answer
C. \left(1 - e^{-\lambda t}\right)\lambda = \dfrac{\ln 2}{t_{1/2}})
Step-by-step solution
The rate of change of the number of active nuclei is the difference between the rate of generation and the rate of decay.
where \dfrac{dN}{R - \lambda N} = dtt = 0tN(0) = 0\int_{0}^{N} \dfrac{dN}{R - \lambda N} = \int_{0}^{t} dt-\dfrac{1}{\lambda} [\ln(R - \lambda N)]_{0}^{N} = t\ln\left(\dfrac{R - \lambda N}{R}\right) = -\lambda t\dfrac{R - \lambda N}{R} = e^{-\lambda t}R - \lambda N = R e^{-\lambda t}N = \dfrac{R}{\lambda} \left(1 - e^{-\lambda t}\right)
where \dfrac{dN}{R - \lambda N} = dtt = 0tN(0) = 0\int_{0}^{N} \dfrac{dN}{R - \lambda N} = \int_{0}^{t} dt-\dfrac{1}{\lambda} [\ln(R - \lambda N)]_{0}^{N} = t\ln\left(\dfrac{R - \lambda N}{R}\right) = -\lambda t\dfrac{R - \lambda N}{R} = e^{-\lambda t}R - \lambda N = R e^{-\lambda t}N = \dfrac{R}{\lambda} \left(1 - e^{-\lambda t}\right)