10mg of 111In is needed for a medical procedure. The sample has to be prepared 3.80 days before the procedure. If 26mg of a 111In sample has to be prepared in advance, what is the half life of 111In?k = ln(2) / t1/2 (Eq 1)ln(m0/mt) = kt (Eq 2)Give your answer in days to two decimal places.
Question
10mg of 111In is needed for a medical procedure. The sample has to be prepared 3.80 days before the procedure. If 26mg of a 111In sample has to be prepared in advance, what is the half life of 111In?k = ln(2) / t1/2 (Eq 1)ln(m0/mt) = kt (Eq 2)Give your answer in days to two decimal places.
Solution
To solve this problem, we need to use the two equations provided.
First, we need to find the decay constant (k) using equation 2.
We know that the initial mass (m0) is 26mg, the final mass (mt) is 10mg, and the time (t) is 3.80 days.
So, we can plug these values into equation 2:
ln(26mg/10mg) = k * 3.80 days
Solving for k gives us:
k = ln(26/10) / 3.80 = 0.1131 days^-1
Now that we have the decay constant, we can find the half-life (t1/2) using equation 1:
t1/2 = ln(2) / k
Plugging in the value for k gives us:
t1/2 = ln(2) / 0.1131 = 6.13 days
So, the half-life of 111In is approximately 6.13 days.
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