SUPPLEMENT: The human body is 0.4% potassium by weight. 0.0117% of all potassium is 4°K. Calculate the rate of decay (dpm) experienced by the average person due to 4°K within their own body. Assume an average weight of 155 lb (70.5 kg). person's weight in grams grams K g grams 40K g moles 4°K moles atoms 4°K atoms 40K beta decay rate constant min- k = ( In 2 ) / ( t,2) decay rate of 40K in body dpm dpm = (k )( atoms 4°K ) This is a much higher decay rate that was produced by the sample of compound in the experiment. But if the probe were held next to your body, practically none would be detected. Explain why.

Chemistry for Engineering Students
4th Edition
ISBN:9781337398909
Author:Lawrence S. Brown, Tom Holme
Publisher:Lawrence S. Brown, Tom Holme
Chapter14: Nuclear Chemistry
Section: Chapter Questions
Problem 14.35PAE
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SUPPLEMENT:
The human body is 0.4% potassium by weight. 0.0117% of all potassium is 4°K. Calculate
the rate of decay (dpm) experienced by the average person due to 40K within their own body.
Assume an average weight of 155 lb (70.5 kg).
person's weight in grams
g
grams K
grams 40K
g
moles 4°K
moles
atoms 40K
atoms
40K beta decay rate constant
min-1
k = ( In 2 ) / ( t,2 )
decay rate of 40K in body
dpm
dpm = (k )( atoms 40K )
%3D
This is a much higher decay rate that was produced by the sample of compound in the experiment.
But if the probe were held next to your body, practically none would be detected. Explain why.
Transcribed Image Text:SUPPLEMENT: The human body is 0.4% potassium by weight. 0.0117% of all potassium is 4°K. Calculate the rate of decay (dpm) experienced by the average person due to 40K within their own body. Assume an average weight of 155 lb (70.5 kg). person's weight in grams g grams K grams 40K g moles 4°K moles atoms 40K atoms 40K beta decay rate constant min-1 k = ( In 2 ) / ( t,2 ) decay rate of 40K in body dpm dpm = (k )( atoms 40K ) %3D This is a much higher decay rate that was produced by the sample of compound in the experiment. But if the probe were held next to your body, practically none would be detected. Explain why.
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