The iodine isotope 13153I is used in hospitals for diagnosis of thyroid function

Question

The iodine isotope 13153I is used in hospitals for diagnosis of thyroid function. 742 g are ingested by a patient. The half-life the iodine isotope 13153I is8.0207 days and mass is 130.906 u.

Part A

Determine the activity immediately after ingestion.

Express your answer using three significant figures.

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Part B

Determine the activity 1.00 h later when the thyroid is being tested.

Express your answer using three significant figures.

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Significant Figures Feedback: Your answer 3.3391012 = 3.339×1012 decays/swas either rounded differently or used a different number of significant figures than required for this part.

Part C

Determine the activity 3.0 months later. Suppose that each month has 30 days.

Express your answer using two significant figures.

1.33•109

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The iodine isotope 13153I is used in hospitals for diagnosis of thyroid function. 742 g are ingested by a patient. The half-life the iodine isotope 13153I is8.0207 days and mass is 130.906 u.

Part A

Determine the activity immediately after ingestion.

Express your answer using three significant figures.

dNdt0 = 3.41×1012   decays/s  

SubmitMy AnswersGive Up

Correct

Part B

Determine the activity 1.00 h later when the thyroid is being tested.

Express your answer using three significant figures.

dNdt = 3.40×1012   decays/s  

SubmitMy AnswersGive Up

Correct

Significant Figures Feedback: Your answer 3.3391012 = 3.339×1012 decays/swas either rounded differently or used a different number of significant figures than required for this part.

Part C

Determine the activity 3.0 months later. Suppose that each month has 30 days.

Express your answer using two significant figures.

dNdt=

1.33•109

decays/s

SubmitMy AnswersGive Up

Incorrect; Try Again; 4 attempts remaining

Not quite. Check through your calculations; you may have made a rounding error or used the wrong number of significant figures.

Explanation / Answer

PART B:

N0 = 3.41×1012 decays/s

After one hour = 3600 seconds, activity is N = N0*exp(-0.693*3600/8.0207*24*3600)

N =3.41×1012 *0.9964 = 3.3977*1012 decays/sec = 3.40*10^12 decays/sec

PARTC:

After 90 DAYS activity is N = N0*exp(-0.693*90*24*3600/8.0207*24*3600)

N =3.41×1012 *4.1964*10^-4= 3.3977*1012 decays/sec = 1.43*10^9 decays/sec

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