First slide

Radio activity

Question

Half life of a radio active material is 5 hours. Then after what time number of undecayed nucleus of a sample of the radio active material will be equal to two times the number of decayed nucleus?

Moderate

A

$5 \times (\frac{\ln 3 - \ln 2}{\ln 3})$
B

$5 (\ln 3 - \ln 2) h r$
C

$5 \times \frac{\ln 3}{\ln 2} h r$
D

$5 \times \frac{\ln 1.5}{\ln 2} h r$

Solution

$\begin{array}{l} N_{0} e^{- λt} = 2 [N_{0} (1 - e^{- λt})] \\ \Rightarrow \frac{3}{2} e^{- λt} = 1 \Rightarrow e^{λt} = \frac{3}{2} \\ \Rightarrow t = \frac{\ln \frac{3}{2}}{λ} = T \times (\frac{\ln 3 - \ln 2}{\ln 2}) \end{array}$

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