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.
Let's work it out. Since the half life is 2 years, the 6 yr span is 3 half life periods. Starting with 10 grams, after the first period (2 years) you will have half of the 10- or 5 grams. After the second period (4 years elapsed) you lose half of what you had- 5 becomes 2.5 After the final period (6 yrs) half of the 2.5 is gone, leaving 1.25 grams.
1 1/4 g APEX
1 g
1 1/4g
1.25
2/1 2 g
You must know the half life of Caesium to calculate this.
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.
11/4 g apex
Approx. 313 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
You must know the half life of Caesium to calculate this.
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.
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.