A) 30 days
B) 50 days
C) 60 days
D) 15 days
Correct Answer: B
Solution :
| \[{{N}_{1}}={{N}_{0}}{{e}^{-\lambda t}}\] \[{{N}_{1}}=\frac{1}{3}{{N}_{0}}\] |
| \[\frac{{{N}_{0}}}{3}={{N}_{0}}{{e}^{-\lambda {{t}_{2}}}}\] |
| \[\Rightarrow \,\,\,\frac{1}{3}={{e}^{-\lambda {{t}^{2}}}}\] ...(i) |
| \[{{N}_{2}}=\frac{2}{3}{{N}_{0}}\] |
| \[\frac{2}{3}{{N}_{0}}={{N}_{0}}{{e}^{-\lambda {{t}_{1}}}}\] |
| \[\Rightarrow \,\,\,\frac{2}{3}={{e}^{-\lambda {{t}_{1}}}}\] ...(ii) |
| Dividing equation (i) by equation (ii) |
| \[\frac{1}{2}={{e}^{-\lambda \,\,({{t}_{2}}-{{t}_{1}})}}\]. |
| \[\lambda ({{t}_{2}}-{{t}_{1}})=In\,2\] |
| \[{{t}_{2}}-{{t}_{1}}=\frac{In\,2}{\lambda }={{T}_{1/2}}=50\] days |
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