1) The predominant isotope of gold, 197 79Au, has an experimentally determined e
ID: 1064163 • Letter: 1
Question
1)
The predominant isotope of gold, 197 79Au, has an experimentally determined exact mass of 196.967 amu. What is the total nuclear binding energy of gold in electronvolts per atom?
Express your answer to three significant figures and include the appropriate units.
2)
A particular smoke detector contains 1.35 Ci of 241Am, with a half-life of 458 years. The isotope is encased in a thin aluminum container. Calculate the mass of 241Am in grams in the detector.
Express your answer numerically in grams.
3)
Neutron activation analysis for a sample of a rock revealed the presence of 5124Cr ,which has a half-life of 27.8 days . Assuming the isotope was freshly separated from its decay products, what is the mass of 5124Cr in a sample emitting 1.00 mCi of radiation?
Express your answer in grams to three significant figures.
Explanation / Answer
1. Given nucleus: 79Au197
No of protons: 79;
No of neutrons: 197-79 = 118
Mass of a proton: 1.00727647 amu
Mass of 79 protons: 79 x 1.007276 = 79.574804 amu
Mass of a neutron: 1.008665 amu
Mass of 79 protons: 118 x 1.008665 =119.02247 amu
Total mass: 119.02247 + 79.574804 =198.597274 amu
Mass Defect: 198.597-196.967 = 1.63 amu
1 amu = 931 MeV
so, 1.63 amu = 931 x 1.63 = 1517.53 MeV
The total nuclear binding energy of gold in electronvolts per atom = 1517.53 MeV = 1.51 GeV
2.
Acitivity of 241Am = 1.35 Ci
t1/2 = 458 years = 458 x 365 x 24 x 60 x 60 = 1.44 x 1010 sec
1 Ci = 3.7 x 1010 dps
so, the activity of 1.35 Ci = 1.35 x 10-6 x 3.7 x 1010 = 4.995 x 104 dps
A = N
= 0.693/t1/2
= 0.693/1.44 x 1010 s = 0.48 x 10-10 s-1
N = A/ = 4.995 x 104 s-1/ 0.48 x 10-10 s-1 = 1.04 x 1014 atoms
N = 1.04 x 1014 atoms
241g of Am contains 6.023 x 1023 atoms
1.04 x 1014 atoms corresponds to = 241 g mol-1x 1.04 x 1014 atoms/ 6.023 x 1023 atoms = 41.61 x 10-9 g
Weight of Am is 41.6 x 10-9 g
3.
Acitivity of 51Cr = 1.00 mCi
t1/2 = 27.8 days = 27.8 x 24 x 60 x 60 = 2.4x 106 sec
1 Ci = 3.7 x 1010 dps
so, the activity of 1.00 mCi = 1.00 x 10-3 x 3.7 x 1010 = 3.7 x 107 dps
A = N
= 0.693/t1/2
= 0.693/2.4x 106 s = 0.28 x 10-6 s-1
N = A/ = 3.7 x 107 s-1/ 0.28 x 10-6 s-1 = 1.321x 1014 atoms
N = 1.321 x 1014 atoms
51g of Cr contains 6.023 x 1023 atoms
1.321 x 1014 atoms = 51 g mol-1x 1.321 x 1014 atoms/ 6.023 x 1023 atoms = 11.18 x 10-9 g
Weight of Cr is 11.2 x 10-9 g .
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