Question #244532

E. Sodium-24 (t - 15 h) is used to study blood circulation. If a patient is injected with an aqueous solution of 2*NaCl that has an activity is 2.5x10° d/s (disintegrations per second). Given hat the disintegration of "Na follows first order kinetics, how much of the activity is present in he patient's blood after 4.0 days


1
Expert's answer
2021-09-30T03:05:02-0400

Na-24 half life = 15h

We know that for first order reaction

t12=0.693Kt\frac{1}{2} = \frac{0.693}{K}

Therefore, K = 0.693t12\frac{0.693}{t\frac{1}{2}}


K= 0.69315h=0.0462h1\frac{0.693}{15h}= 0.0462h^{-1}


At starting, (Activity)0 = 2.5x109d/s


At 4days, 4×24hr= 96hr

96×2.5×109

= 2.4×1011 d/s


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