The mass is 1,075 g.
The half-life of thorium-234 is about 24 days. Therefore, it would take approximately 96 days for one-sixteenth of the original 54.2 g sample of thorium-234 to remain.
A half-life of a radioisotope is the time required for half of a sample to decay. In this case, a 50-g sample becoming 25 g after 18 days indicates that the half-life of the radioisotope is 18 days, as the sample has decreased to half its original amount in that time.
After 4 days, only 1/16 of the original amount of gold (198/16 = 12.375) will remain.
After 8.1 days, three half-lives have passed (8.1 days / 2.7 days = 3). With each half-life, the number of atoms is halved. Therefore, starting with 800 atoms, after three half-lives there would be 800 / 2 / 2 / 2 = 100 atoms remaining.
After 16.14 days, which is two half-lives of iodine-131 (since 16.14 days / 8.07 days = 2), the amount of iodine-131 remaining would be (200 , \text{g} \times \left(\frac{1}{2}\right)^2 = 50 , \text{g}). Therefore, the amount of iodine-131 that has decayed to xenon-131 would be (200 , \text{g} - 50 , \text{g} = 150 , \text{g}). Thus, after 16.14 days, there would be 150 grams of xenon-131 formed.
The mass is 1,075 g.
The answer depends on 3240 WHAT: seconds, days, years?
The half-life of thorium-234 is about 24 days. Therefore, it would take approximately 96 days for one-sixteenth of the original 54.2 g sample of thorium-234 to remain.
After 48,2 days the amount of Th-234 will be 25 g.
After 76 seconds, half of the radium-222 would have decayed (its half-life is about 3.8 days). Therefore, the quantity of radium-222 remaining in the 12-gram sample would be 6 grams.
At 8.1 days, 400 atoms of Au198 would remain in the sample. This is because after 8.1 days, two half-lives of Au198 have passed, reducing the initial 800 atoms to 400.
Thorium-234 has a half-life of 24.1 days. How much of a 100-g sample of thorium-234 will be unchanged after 48.2 days?
79 grams <><><><><> AT = A0 2 (-T/H) AT = 250 2(-40/24.1) AT = ~79
After 32 days, approximately 5 milligrams of the 80-milligram sample of Iodine-131 would be left. Iodine-131 has a half-life of about 8 days, so after each 8-day period, half of the remaining sample will decay.
After 42 days there would only be 1.00mg of Phosphorus-32 left in the organisms system
After 28 days, two half-lives would have passed (14 days x 2). This means the initial amount of phosphorus-32 would be reduced by three-quarters (1/2 x 1/2 = 1/4). Therefore, 6 mg (24 mg x 1/4) of phosphorus-32 will remain after 28 days.
After 9 days, the population of ladybugs would double every 3 days, so it would double 3 times. 2^3 = 8. Therefore, the population at the end of 9 days would be 30 ladybugs x 8 = 240 ladybugs.