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a sample of 90Ic (T1/2 = 6h) is assayed to you an activity of 9mci. it is administered by intravenous injection to a patient suspected by having carcinoma of the lungs 49 hour later. what is the activity in Bq as that time?

User Blazemonger
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19 votes

Answer: I got you <3

Step-by-step explanation:

To determine the activity of the sample in Becquerels (Bq) 49 hours after it was administered, you can use the following formula:

A(t) = A(0) * e^(-lambda * t)

Where:

A(t) is the activity at time t

A(0) is the initial activity

lambda is the decay constant, which is equal to ln(2) / T1/2

t is the elapsed time

Plugging in the values provided in the question, we get:

A(49 h) = 9 mci * e^(- (ln(2) / 6 h) * 49 h)

= 9 mci * e^(- (0.1155) * 49 h)

= 9 mci * e^(-5.6475)

= 9 mci * 0.0023

= 0.0207 mci

Converting mci to Bq:

1 mci = 37 GBq

0.0207 mci = (0.0207 mci) * (37 GBq / mci)

= 0.7649 GBq

So the activity of the sample in Bq 49 hours after it was administered is approximately 0.7649 GBq.

User Imre L
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