The half-life of the Nobelium isotope No^257 is about 23 seconds. 644 seconds (or 28 half-life periods) after the isotope was released there were 20 grams remaining. The number of grans of No^257 after h half life periods is M= 20 * (1/2)^h where h = 0 corresponds to 644 seconds after the isotope was released. How much No^257 initially released? Answer in units of g.
20=A(.5)^-28
A=20(2)^28
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20*2^28=
To find out how much No^257 was initially released, we can use the given formula for M, which represents the number of grams remaining after h half-life periods:
M = 20 * (1/2)^h, where h = 0 corresponds to 644 seconds.
We are given that after 28 half-life periods, there were 20 grams remaining. Therefore, we need to find the initial amount.
Let's substitute the values into the formula and solve for M when h = 28:
M = 20 * (1/2)^h
M = 20 * (1/2)^28
Now, we need to find the initial amount, which we can denote as M_initial. We can rearrange the formula as follows:
M_initial = M / (1/2)^h
M_initial = 20 * (1/2)^28 / (1/2)^28
M_initial = 20 grams
Therefore, the initial amount of No^257 released was 20 grams.