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A samll amount of solution containing Na^(24) radio nuclide with activity A = 2xx10^(3) dps was administeredinto blood of a patientin a hospital. Afer 5 hour a sample of the blooddrawn out form the patientshowed an activity of 16 dpm per cc t_(1//2) for Na^(24) = 15 hr. Find: (a) Volume of the blood in the patient. (b) Activity fo blood sample drawn after a further time fo 5 hr.

Answer» <html><body><p></p>Solution :Let `V mL` blood is present in patient. <br/> (a) `r_(0)` pf `Na^(24) = 2xx10^(3) <a href="https://interviewquestions.tuteehub.com/tag/dps-432950" style="font-weight:bold;" target="_blank" title="Click to know more about DPS">DPS</a> = 2xx10^(3) xx 60 dp m` <br/> `= 120xx10^(3) dp m` for `V mL` blood <br/> `r` of `Na^(24) = <a href="https://interviewquestions.tuteehub.com/tag/16-276476" style="font-weight:bold;" target="_blank" title="Click to know more about 16">16</a> dp m//mL` at `t = 5 <a href="https://interviewquestions.tuteehub.com/tag/hr-25457" style="font-weight:bold;" target="_blank" title="Click to know more about HR">HR</a>`<br/> `= 16xx V dp m//V mL`<br/> `:' (r_(0))/(r) = (N_(0))/(N)` <br/> `:. (N_(0))/(N) = (120xx10^(3))/(16V)` <br/> `:. t = (2.303)/(lambda) log_(10) ((N_(0)))/(N)` <br/> `5 = (2.303xx15)/(0.693) log_(10)((120xx10^(3)))/(16V)` <br/> `:. V = 5.95 xx10^(3) mL` <br/> (b) Activity of blood sample after`5 hr` more i.e., `t = 10 hrt` <br/> `t = (2.303)/(lambda) log_(10) (N_(0))/(N)` <br/> `10 = (2.303xx15)/(0.693) log_(10) (120xx10^(3))/(A)` <br/> `:. A = 75.6xx10^(3) dp m` per `5.95 xx10^(3)mL` <br/> `= (75.6xx10^(3))/(5.95xx10^(3)) = 12.71` dpm per `mL`. <br/> `= 0.2118` dpm per `mL`.</body></html>


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