Answer: 7.82 days
Solution:
N=N0×(1/2)tT1/2N=N_{0}×(1/2)^{\frac{t}{T_{1/2}}}N=N0×(1/2)T1/2t
lnNN0=−tT1/2×ln2\ln{\frac{N}{N_{0}}}=-{\frac{t}{T_{1/2}}}×\ln2lnN0N=−T1/2t×ln2
t=−lnNN0×T1/2ln2=−lnCC0×T1/2ln2t=-\frac{\ln{\frac{N}{N_{0}}}×{T_{1/2}}}{\ln2}=-\frac{\ln{\frac{C}{C_{0}}}×{T_{1/2}}}{\ln2}t=−ln2lnN0N×T1/2=−ln2lnC0C×T1/2
t=−ln300 μmol/L1250 μmol/L×3.8 daysln2=7.82 dayst=-\frac{\ln{\frac{300\,\mu mol/L}{1250\,\mu mol/L}}×{3.8\,days}}{\ln2}=7.82\,dayst=−ln2ln1250μmol/L300μmol/L×3.8days=7.82days
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