Reference no: EM13954275
1. A wooden archeological artifact is reported to be 2000 years old on the basis of carbon-14 dating. What is the fraction of carbon-14 found in the specimen to that found in new wood? The half-life of carbon-14 is 5700 years.
a. 1.0
b. 0.0
c. 0.784
d. 0.627
2. How many protons and how many neutrons are needed to make the following elements: 200Hg80, 16O8, and 232Pb82?
a. 80 protons and 120 neutrons 8 protons and 8 neutrons 90 protons and 142 neutrons
b. 120 protons and 80 neutrons 16 protons and 0 neutrons 142 protons and 90 neutrons
c. 75 protons and 125 neutrons 6 protons and 10 neutrons 80 protons and 152 neutrons
d. 70 protons and 130 neutrons 9 protons and 7 neutrons 100 protons and 132 neutrons
3. Natural copper occurs as a mixture of two isotopes: The isotope 63Cu29 has a relative abundance of 69%, while the isotope 65Cu29 has a relative abundance of 31%. What is the atomic mass of natural copper?
a. 63
b. 65
c. 63.6
d. 64.3
4. What is the approximate mass number of a nucleus whose radius is measured to be 6.0x10^-15 m?
a. 110
b. 90
c. 82
d. 125
5. Calculate the binding energy of 9Be4 using 9.012182 Mass(u), 8394.796 Mass(MeV) 14.96509(10^-27 kg)
a. 58.166 MeV
b. 26.124 MeV
c. 7.654 MeV
d. 120.15 MeV
6. Identify the nucleus designated by X in each of the following a)226Ra88 -----> X + alpha b) 233Pa91 -----> X + beta(-) c)59Fe26 -------> X + gamma
a. a)212Rn86 b) 239U92 c)57Fe26
b. a)232Rn86 b) 231U92 c)58Fe26
c. a)230Rn86 b) 234U92 c)60Fe26
d. a)222Rn86 b) 233U92 c)59Fe26
7. The isotope 234U92 undergoes five successive alpha decays. Identify the daughter nucleus at each step.
a. 230Th90 --> 226Ra88 --> 222Rn86 --> 218Po84 --> 214Pb82
b. 232Th90 --> 228Ra88 --> 224Rn86 --> 220Po84 --> 216Pb82
c. 228Th90 --> 224Ra88 --> 220Rn86 --> 216Po84 --> 212Pb82
d. 233Th90 --> 229Ra88 -->225Rn86 --->218Po84 --> 217Pb82
8. Calculate the approximate Q value of the decay 218Po84 ----> 214Pb82 + alpha given the measured kinetic energy of the alpha is 5.998 MeV
a. 5.12 MeV
b. 6.11 MeV
c. 7.84 MeV
d. 1.75 MeV
9. a) Determine the kinetic energy available to products of alpha decay of 235U92.
b) Also find the kinetic energy of the alpha particle. The observed masses are m(235U92) = 235.04392 u and m(231Th90) = 231.03630 u.
a. 4.68 MeV and 4.60 MeV
b. 3.68 MeV and 3.98 MeV
c. 2.19 MeV and 2.00 MeV
d. 1.76 MeV and 1.50 MeV
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: Determine the kinetic energy available to products of alpha decay of 235U92. b) Also find the kinetic energy of the alpha particle. The observed masses are m(235U92) = 235.04392 u and m(231Th90) = 231.03630 u.
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