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.

asked by J

9 years ago

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1 answer

20=A(.5)^-28
A=20(2)^28
put this in your google search engine:
20*2^28=

answered by bobpursley

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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.

1 answer

20=A(.5)^-28
A=20(2)^28
put this in your google search engine:
20*2^28=

Ask a New Question or answer this question.

bobpursley · Answer

20=A(.5)^-28 A=20(2)^28 put this in your google search engine: 20*2^28=