Americium-244 is a rare isotope of Americium. What is the mass defect of Americi

Question

Americium-244 is a rare isotope of Americium. What is the mass defect of Americium-244? Use the following values for atomic and neutron masses when calculating your answer: Am: 244.004279u H: 1.007825u n: 1.008665 u Check Americium-244 undergoes beta (B-) decay. Select all of the products of this decay from the list below Select one or more Check The half life of Americium-244 is 10.1 hours. After time 21.9 hours, how much of Americium-244 would be left in a sample that initially contained 12.7 g of Americium- 244? heck

Explanation / Answer

A) Mass defect M = (mass of Am-244 - mass of H-1 x number of proton - mass of n-0 x number of neutron)

M = 244.064279 - 1.007825 x 95 - 1.008665 x (244-95) = 1.97018

B) The reaction for the beta decay will be 244Am95 -> 244Cm96 + e- + e. Accordingly, the product is mentioned on the right of the nuclear reaction.

C) T1/2 = 10.1 hours. Initial mass mi= 12.7 g. = Log(2)/T1/2 = 0.06863 /s.

At T = 21.9 hours, the Americium-22 left mf = mi exp(-T)

mf = 12.7 exp(-0.06863 x 21.9) = 2.8253 g.

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