A) 20 min
B) 30 min
C) 40 min
D) 25 min
Correct Answer: C
Solution :
[c] : According to radioactive decay, \[N={{N}_{0}}{{e}^{-\lambda t}}\]where, \[{{N}_{0}}=\]Number of radioactive nuclei present in the sample at t = 0 N = Number of radioactive nuclei left undecayed after time t \[\lambda =\]decay constant For 20% decay \[\frac{80{{N}_{0}}}{100}={{N}_{0}}{{e}^{-\lambda {{t}_{1}}}}\] ...(i) For 80% decay \[\frac{20{{N}_{0}}}{100}={{N}_{0}}{{e}^{-\lambda {{t}_{2}}}}\] ....(ii) Dividing equation (i) by (ii), we get \[4={{e}^{-\lambda ({{t}_{1}}-{{t}_{2}})}}\] \[\Rightarrow 4={{e}^{\lambda ({{t}_{2}}-{{t}_{1}})}}\] Taking natural logarithms of both sides of above equation, we get \[\ln 4=\lambda ({{t}_{2}}-{{t}_{1}})\] \[2\ln 2=\frac{\ln 2}{{{T}_{1/2}}}({{t}_{2}}-{{t}_{1}})\] \[{{t}_{2}}-{{t}_{1}}=2\times {{T}_{1/2}}=2\times 20\min =40\min \]
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