Problem S: Planets Beyond the Solar System. On October 15, 2001, a planet was discovered orbiting around the star HD 68988. Its orbital distance was measured to be 1.05 x 10¹⁰ m from the center of the star, and its orbital period was estimated at 6.3 days = 5.44 x 105 sec. What is the mass of HD 68988? Express your answer in kilograms and in terms of our sun's mass (Msun = 1.99 x 1030 kg) Answer: M = 2.31 x 10³⁰ kg = 1.16 Msun-

Problem S: Planets Beyond the Solar System. On October 15, 2001, a planet was discovered orbiting around the star HD 68988. Its orbital distance was measured to be 1.05 x 10¹⁰ m from the center of the star, and its orbital period was estimated at 6.3 days = 5.44 x 105 sec. What is the mass of HD 68988? Express your answer in kilograms and in terms of our sun's mass (Msun = 1.99 x 1030 kg) Answer: M = 2.31 x 10³⁰ kg = 1.16 Msun-

College Physics

10th Edition

ISBN:9781285737027

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter29: Nuclear Physics

Section: Chapter Questions

Problem 2P

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Similar questions

What is the mass of 60Co in a cancer therapy transillumination unit containing 5.00 kCi of 60Co?
(a) If the average molecular mass of compounds in food is 50.0 g, how many molecules are mere in 1.00 kg at food? (b) How many ion pairs are created in 1.00 kg of food, if it is exposed to 1000 Sv and it takes 32.0 eV to create an ion pair? (c) Find the ratio of ion pairs to molecules. (d) If these ion pairs recombine into a distribution of 2000 new compounds, how many parts per billion is each?
Data from the appendices and the periodic table may be needed for these problems. A 5000Ci 60Co source used for cancer therapy is considered too weak to be useful when its activity falls to 3500 Ci. How long after its manufacture does this happen?
The Galileo space probe was launched on its long journey past Venus and Earth in 1989, with an ultimate goal of Jupiter. Its power source is 11.0 kg of 238Pu, a by-product of nuclear weapons plutonium production. Electrical energy is generated thermoelectrically from the heat produced when the 5.59-MeV a panicles emitted in each decay crash to a halt inside the plutonium and its shielding. The half-life of 238Pu is 87.7 years. What was the original activity of the 238Pu in becquerels? What power was emitted in kilowatts? What power was emitted 12.0 y after launch? You may neglect any extra energy from daughter nuclides and any losses from escaping rays.
(a) How many 239Pu nuclei must fission to produce a 20.0kT yield, assuming 200 MeV per fission? (b) What is the mass of this much 239Pu?
What is the dose in mSv for: (a) a 0.1 Gy xray? (b) 2.5 mGy of neutron exposure to the eye? (c) 1.5 mGy of exposure?
(a) Show that if you assume the average nucleus is spherical with a radius r=r0A1/3, and with a mass at A u, then its density is independent at A. (b) Calculate that density in u/fm3 and kg/m3, and compare your results with those found in Example 31.1 for 56Fe.
A neutron generator uses an (source, such as radium, to bombard beryllium, inducing the reaction 4He+9Be12C+n. Such neutron sources are called RaBe sources, 0r PuBe sources if they use plutonium to get the (s. Calculate the energy output of the reaction in MeV.
The purpose of producing 99Mo (usually by neutron activation of natural molybdenum, as in the preceding problem) is to produce 99mTc. Using the rules, verily that the decay of 99Mo produces 99mTc. (Most 99mTc nuclei produced in this decay are left in a metastable excited state denoted 99mTc.)
Enter the correct nuclide symbol in each open tan rectangle in Figure P43.25, which shows the sequences of decays in the natural radioactive series starting with the long-lived isotope uranium-235 and ending with the stable nucleus lead-207. Figure P43.25
Data from the appendices and the periodic table may be needed for these problems. 44. An old camp?re is uncovered during an archaeological dig. Its charcoal is found to contain less than 1/1000 the normal amount of 14C. Estimate the minimum age at the charcoal, noting that 210 = 1024.