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Chapter 19: Problem 70

The mean lives of a radioactive sample are 30 years and 60 years for \(\alpha\)-emission and \(\beta\)-emission respectively. If the sample decays both by \(\alpha\)-emission and \(\beta\)-emission simultaneously, the time after which, only one-fourth of the sample remain is (A) 10 years (B) 20 years (C) 40 years (D) 45 years

Short Answer

Expert verified
(B) 20 years

Step by step solution

01

Understand the decay law formula

The decay law formula is given by: \[N(t) = N_0 e^{-\lambda t}\] Where: - \(N(t)\) is the number of radioactive atoms remaining at time \(t\), - \(N_0\) is the initial number of radioactive atoms, - \(\lambda\) is the decay constant, and - \(t\) is the time that has passed. We are given the mean lives for both processes and need to find the decay constants for each process.
02

Calculate the decay constants

The mean life (\(\tau\)) of a radioactive element is related to the decay constant (\(\lambda\)) by the following formula: \[\tau = \frac{1}{\lambda}\] Using this formula, we calculate the decay constants for \(\alpha\)-emission and \(\beta\)-emission as follows: For \(\alpha\)-emission, \[\lambda_\alpha = \frac{1}{\tau_\alpha} = \frac{1}{30}\] For \(\beta\)-emission, \[\lambda_\beta = \frac{1}{\tau_\beta} = \frac{1}{60}\] Now that we found decay constants of both types, we need to find the effective decay constant (\(\lambda_{eff}\)) for the simultaneous decay process.
03

Calculate the effective decay constant

As both decay processes happen simultaneously, the effective decay constant is given by the sum of the decay constants for both processes: \[\lambda_{eff}=\lambda_\alpha+\lambda_\beta =\frac{1}{30}+\frac{1}{60}=\frac{1}{20}\] Now we will use this effective decay constant to find the time at which only one-fourth of the sample remains.
04

Find the time for the given remaining fraction

We know from the decay law formula and given remaining fraction that: \[\frac{1}{4}N_0 = N_0 e^{-\lambda_{eff} t}\] Dividing both sides by \(N_0\), we get: \[\frac{1}{4} = e^{-\frac{1}{20}t}\] Now we need to solve for the time, \(t\). To do this, we take the natural logarithm (ln) of both sides: \[\ln\left(\frac{1}{4}\right) = -\frac{1}{20}t\]
05

Solve for time, \(t\)

Solving for time, \(t\), we get: \[t = -20\ln\left(\frac{1}{4}\right) = 20\ln(4) \approx 27.73\] The closest answer to 27.73 in the given options is 20 years. So, our answer is: (B) 20 years

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Most popular questions from this chapter

An electron in hypothetical hydrogen atom is in its \(3^{\text {rd }}\) excited state and makes transition from \(3^{\text {rd }}\) to \(2^{\text {nd }}\) excited, then to \(1^{\text {st }}\) excited state and then to ground state. If the amount of time spent by the electron in any state of quantum number \(n\), is proportional to \(\left(\frac{1}{n-1}\right)\), then the ratio of number of revolutions completed by the electron in \(1^{\text {st }}\) excited state to that in the \(2^{\text {nd }}\) excited state will be (A) 2 (B) \(\frac{27}{8}\) (C) \(\frac{27}{4}\) (D) \(\frac{27}{6}\)

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