You must know the half life of Caesium to calculate this.
11/4 g apex
2 1/2g
Since the half-life of cesium-137 is about 30 years, 3 half-lives would have passed in 90 years. The first half-life would leave .5 mg of cesium-137. The second would leave .25 mg, and the third half-life would leave .175 mg of cesium-137.
After 10740 years, half of the sample would have decayed, so there would be 200 atoms left. If the original sample had 400 atoms, then there would be 200 atoms left in the sample after 10740 years.
The formula for finding the amount of radio active substance that remainsafter decomposition is A = N(1/2) ^y, where A is the mass remaining, N is the original mass of a radioactive sample and y is the number of half life. In this case, N = 10 g and y = 6/2 = 3. So A is given by 10 x (1/2)^3, which is 10 x 1/2 x1/2 x 1/2 = 1.25 grams.
5g would remain
I suppose that you think to the radioactive isotope Cs-17; After 4 years remain 9,122 g.
2 1/2 g
2 1/2 g
2 1/2 g
2 1/2 g
2 1/2 g
1 1/4 g (apex)or 1.25 g
11/4 g apex
2 1/2g
As you did not specify an isotope of cesium, I will assume you meant natural cesium. Natural cesium is not radioactive so it does not decay. There will always be the same 10 g of cesium, no matter how long you wait.
Since the half-life of cesium-137 is about 30 years, 3 half-lives would have passed in 90 years. The first half-life would leave .5 mg of cesium-137. The second would leave .25 mg, and the third half-life would leave .175 mg of cesium-137.