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 μCi 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.

Answer 1

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 μCi

t_1/2 = 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 μCi = 1.35 x 10^-6 x 3.7 x 10¹⁰ = 4.995 x 10⁴ dps

A = Nλ

λ = 0.693/t_1/2

λ = 0.693/1.44 x 10¹⁰ s = 0.48 x 10^-10 s^-1

N = A/λ = 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

t_1/2 = 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 = Nλ

λ = 0.693/t_1/2

λ = 0.693/2.4x 10⁶ s = 0.28 x 10^-6 s^-1

N = A/λ = 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 ^.