A) 40 minutes
B) 10 minutes
C) 30 minutes
D) 25 minutes
Correct Answer: A
Solution :
: Let in time \[{{t}_{1}}\], 50% of the substance decay and in time \[{{t}_{2}}\] 87.5% of the substance decay. Then in time \[{{t}_{1}}\], 50% of the substance left undecayed and in time \[{{t}_{2}}\] 12.5% of the substance left undecayed. According to radioactive decay law \[N={{N}_{0}}{{e}^{-\lambda t}}\]where \[\lambda \] is the decay constant. or \[\frac{N}{{{N}_{0}}}={{e}^{-\lambda t}}\] \[\therefore \] \[\frac{50}{100}={{e}^{-\lambda {{t}_{1}}}}\] or \[\frac{1}{2}={{e}^{-\lambda {{t}_{1}}}}\] ?.(i) and \[\frac{12.5}{100}={{e}^{-\lambda {{t}_{2}}}}\] or \[\frac{1}{8}={{e}^{-\lambda {{t}_{2}}}}\] ??(ii) Dividing eqn. (ii) by eqn. (i), we get \[\frac{\frac{1}{8}}{\frac{1}{2}}=\frac{{{e}^{-\lambda {{t}_{2}}}}}{{{e}^{-\lambda {{t}_{1}}}}}\] or \[\frac{1}{4}={{e}^{-\lambda ({{t}_{2}}-{{t}_{1}})}}\] or \[\lambda ({{t}_{2}}-{{t}_{1}})=ln4\] \[{{t}_{2}}-{{t}_{1}}=\frac{1n4}{\lambda }=\frac{21n2}{\lambda }\](\[\because \] \[1n4=21n2\]) \[=\frac{21n2}{\left( \frac{1n2}{{{T}_{1/2}}} \right)}=2{{T}_{1/2}}\] \[\left( \because \lambda =\frac{1n2}{{{T}_{1/2}}} \right)\] Here, \[{{T}_{1/2}}=20\] minutes \[\therefore \] \[{{t}_{2}}-{{t}_{1}}=2\] (20 minutes) = 40 minutes
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