The equation for half-life is ...
AT = A0 2 (-T/H)
... where A0 is the starting activity, AT is the activity at some time T, and H is the half-life, in units of T.
55134Cs has a half-life of 2.0652 years. Plugging in the known values, we get ...
AT = 5.8 2 (-11.5/2.0652)
AT = 5.8 2 -5.5685
AT = 0.12222
5g would remain
I suppose that you think to the radioactive isotope Cs-17; After 4 years remain 9,122 g.
The half-life of cesium-137 is approximately 30.1 years, not 2 years. After one half-life, 5 G of the original 10 G sample would remain. After two half-lives (about 60.2 years), 2.5 G would remain, and so on. If you meant a hypothetical isotope with a 2-year half-life, after 2 years, 5 G would remain, and after 4 years, 2.5 G would remain.
2 1/2 g
2 1/2 g
5g would remain
I suppose that you think to the radioactive isotope Cs-17; After 4 years remain 9,122 g.
The half-life of cesium-137 is approximately 30.1 years, not 2 years. After one half-life, 5 G of the original 10 G sample would remain. After two half-lives (about 60.2 years), 2.5 G would remain, and so on. If you meant a hypothetical isotope with a 2-year half-life, after 2 years, 5 G would remain, and after 4 years, 2.5 G would remain.
2 1/2 g
2 1/2 g
2 1/2 g
2 1/2 g
2 1/2 g
After 6 years, approximately 5 grams of cesium-137 would remain from a 10 g sample due to its half-life of around 30 years. This decay is exponential, with about half of the original sample decaying every 30 years.
1 1/4 g (apex)or 1.25 g
11/4 g apex
2 1/2g