JEE Mains · Physics · STD 12 - 13. Nuclei
A radioactive sample has an average life of \(30\, {ms}\) and is decaying. A capacitor of capacitance \(200\, \mu\, {F}\) is first charged and later connected with resistor \(^{\prime}{R}^{\prime}\). If the ratio of charge on capacitor to the activity of radioactive sample is fixed with respect to time then the value of \(^{\prime}R^{\prime}\) should be \(....\,\Omega\)
- A \(100\)
- B \(200\)
- C \(150\)
- D \(250\)
Answer & Solution
Correct Answer
(C) \(150\)
Step-by-step Solution
Detailed explanation
\(q=q_{0} e^{-\frac{t}{R C}}\) where, \(R C\) is the time constant of \(R C\) - circuit. At time \(t\), activity \(A\) is given by \(A=A_{0} e ^{-\lambda t} \) where, \(A_{0}=\) initial activity and \(\quad \lambda=\) decay constant. On dividing Eqs. (i) and (ii), we get…
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