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The total decay rate R of a sample is related to the decay rate R0\text R_0R0​​ at t = 0 and the disintegration constant or decay constant λ as:
Question

The total decay rate R of a sample is related to the decay rate R0\text R_0​ at t = 0 and the disintegration constant or decay constant λ as:

A.

R=R0 e(λ)t\text{R}=\text{R}_0\ e^{-(λ)t}​​

B.

R=R0 e(2λ)t\text{R}=\text{R}_0\ e^{-(2λ)t}​​

C.

R=R0 e(λ2)t\text{R}=\text{R}_0\ e^{-(\frac{λ}{2})t}​​

D.

R=R0 e(λ4)t\text{R}=\text{R}_0\ e^{-(\frac{λ}{4})t}​​

Correct option is A

The rate of change of radioactive nuclei is proportional to the number of undecayed nuclei:dRdt=λRWhere:R = decay rate at time tλ = decay constantR0 = initial decay rate at t=0\text{The rate of change of radioactive nuclei is proportional to the number of undecayed nuclei:} \\\frac{dR}{dt} = -\lambda R \\[10pt]\text{Where:} \\R \text{ = decay rate at time } t \\\lambda \text{ = decay constant} \\R_0 \text{ = initial decay rate at } t = 0

1R dR=λdtlnR=λt+CWhere C is the integration constant.\int \frac{1}{R} \, dR = -\lambda \int dt \\[5pt]\ln R = -\lambda t + C \\[10pt]\text{Where } C \text{ is the integration constant.}

At t=0, R=R0lnR0=ClnR=λt+lnR0ln(RR0)=λtR=R0eλt\text{At } t = 0, \, R = R_0 \\[5pt]\ln R_0 = C \\[10pt]\ln R = -\lambda t + \ln R_0 \\[5pt]\ln \left( \frac{R}{R_0} \right) = -\lambda t \\[5pt]R = R_0 e^{-\lambda t}​​​​

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