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abri
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I'm taking an introduction course for simple physics. All problems are solved with simple single variable analysis. My problem is that this is an introduction where you don't need any advanced math to solve the problems, but I can't remember how to solve the last equation. Would you mind helping me with this?
1. Homework Statement
Cobalt-60 (60Co) is often used in nuclear medicine. It has a half life of 1925 days (5,27 years).
A) Find the activity in a sample of 1,0 µg of cobalt-60.
B) An old sample originally contained 10 µg of cobalt-60 and has the activity of 5,0 MBq, how old is it?
N=N0*e^(−λt)
I've found the activity in the first sample to be 41,83 MBq.
T½ in sec = 1925 days * 24h * 60 min * 60 sec = 166 320 000 sec
Mean lifetime τ = T½ / ln2 = 239 965 373 sec
Decay rate λ = 1/τ = 4,167 * 10 ^-9 / sec
Cobalt-60 has an atomic weight of 60 u, meaning one mole weighs 60 grams.
So out of a 1,0 µg sample this mass will decay each second:
(4,167 * 10 ^-9 / sec) x (1 * 10^-6 mass in gram) = 4,167 * 10^-15 grams / second
This equates to:
(4,167 * 10^-15 grams / sec) x (1.0 mole / 59,93 g) x (6,022 x 10^23 atoms / mole) = 41,87*10^6 atoms/second
Activity in a 1 µg sample is 41,87 MBq
That means the activity in a 10 µg sample at t=0 is 418,3 MBq.
The activity is directly proportional to the number of nuclei left in the sample. This means we can use the formula:
N=N0*e^(−λt)
(Activity at time t) = (Activity at time 0) * e^(−λt)
This is where I get stuck, since it's an introduction you should be able to solve it with simple math but I can't figure out how to solve for t in the equation above. Any and all help would be appreciated.
1. Homework Statement
Cobalt-60 (60Co) is often used in nuclear medicine. It has a half life of 1925 days (5,27 years).
A) Find the activity in a sample of 1,0 µg of cobalt-60.
B) An old sample originally contained 10 µg of cobalt-60 and has the activity of 5,0 MBq, how old is it?
Homework Equations
N=N0*e^(−λt)
The Attempt at a Solution
I've found the activity in the first sample to be 41,83 MBq.
T½ in sec = 1925 days * 24h * 60 min * 60 sec = 166 320 000 sec
Mean lifetime τ = T½ / ln2 = 239 965 373 sec
Decay rate λ = 1/τ = 4,167 * 10 ^-9 / sec
Cobalt-60 has an atomic weight of 60 u, meaning one mole weighs 60 grams.
So out of a 1,0 µg sample this mass will decay each second:
(4,167 * 10 ^-9 / sec) x (1 * 10^-6 mass in gram) = 4,167 * 10^-15 grams / second
This equates to:
(4,167 * 10^-15 grams / sec) x (1.0 mole / 59,93 g) x (6,022 x 10^23 atoms / mole) = 41,87*10^6 atoms/second
Activity in a 1 µg sample is 41,87 MBq
That means the activity in a 10 µg sample at t=0 is 418,3 MBq.
The activity is directly proportional to the number of nuclei left in the sample. This means we can use the formula:
N=N0*e^(−λt)
(Activity at time t) = (Activity at time 0) * e^(−λt)
This is where I get stuck, since it's an introduction you should be able to solve it with simple math but I can't figure out how to solve for t in the equation above. Any and all help would be appreciated.