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  • The half life of radioactive nucleus is 50 days. The time interval between the time <img alt="t_2" src="/latex-im

The half life of radioactive nucleus is 50 days. The time interval \left(t_2-t_1\right) between the time t_2 when \frac{2}{3} of it has decayed and the time t_1 when \frac{1}{3} of it had decayed is:

Option: 1

30 days


Option: 2

50 days


Option: 3

60 days


Option: 4

15 days


Answers (1)

best_answer

According to radioactive decay law \mathrm{N}=\mathrm{N}_0 \mathrm{e}^{-\lambda \mathrm{t}}

Where \mathrm{N}_0=  Number of radioactive nuclei at time  \mathrm{t}=0

\mathrm{N}= Number of radioactive nuclei left undecayed at any time \mathrm{t}

\lambda= decay constant 

At time  t_2, \frac{2}{3} of the sample had decayed 

\therefore \mathrm{N}=\frac{1}{3} \mathrm{~N}_0 \quad \text { or } \quad \frac{1}{3} \mathrm{~N}_0=\mathrm{N}_0 \mathrm{e}^{-\lambda \mathrm{t}_2}                       (i)

At time  t_1, \frac{1}{3} of the sample had decayed, 

\therefore \mathrm{N}=\frac{2}{3} \mathrm{~N}_0 \quad \text { or } \quad \frac{2}{3} \mathrm{~N}_0=\mathrm{N}_0 \mathrm{e}^{-\lambda \mathrm{t}_1}                     (ii)

Divide (i) by (ii), we get

\frac{1}{2}=\frac{e^{-\lambda t_2}}{e^{-\lambda t_1}} \quad \text { or } \quad \frac{1}{2}=e^{-\lambda\left(t_2-t_1\right)} \quad \text { or } \quad \lambda\left(t_2-t_1\right)=\ln 2

\mathrm{t}_2-\mathrm{t}_1=\frac{\ln 2}{\lambda}=\frac{\ln 2}{\left(\frac{\ln 2}{\mathrm{~T}_{1 / 2}}\right)} \quad\left(\because \lambda=\frac{\ln 2}{\mathrm{~T}_{1 / 2}}\right)

\mathrm{T}_{1 / 2}=50 \text { days }

Posted by

Ritika Jonwal

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