To solve this problem, we can use the formula for exponential decay:
A = Ao(1 - r)^t
Where:
A = final mass after time t
Ao = initial mass
r = decay rate per second
t = time in seconds
We know that the decay rate is 4.9% or 0.049, and the initial mass is 100 mg. Substituting these values into the formula, we get:
A = 100(1 - 0.049)^t
Now, we need to find the half-life of Beryllium-11. The half-life is the time it takes for half of the radioactive substance to decay.
Let's assume that after a certain time t, the mass A decreases to half of its initial value Ao:
A = Ao/2
Substituting this into the equation above, we get:
Ao/2 = Ao(1 - 0.049)^t
Dividing both sides by Ao, we get:
1/2 = (1 - 0.049)^t
Taking the logarithm of both sides, we get:
log(1/2) = log[(1 - 0.049)^t]
Using the logarithm property log(a^b) = b*log(a), we can rewrite this as:
log(1/2) = t*log(1 - 0.049)
Solving for t, we get:
t = log(1/2) / log(1 - 0.049)
Using a calculator, we find:
t ≈ 14.44 seconds
Therefore, the half-life of Beryllium-11 is approximately 14.44 seconds.
A 100mg sample of Beryillium 11 decays by 4.9% of its previous mass every second. Determine the half-life of Beryillium-11.
[Hint: the equation A=Ao(1-r)^t may be a useful starting point.]
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