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The critical density of the universe is $5.8 \times 10^{-27} \mathrm{kg} / \mathrm{m}^{3}$ . (a) Assuming that the universe is all hydrogen, express the critical density in the number of $\mathrm{H}$ atoms per cubic meter. (b) If the density of the universe is equal to the critical density, how many atoms, on the average, would you expect to find in a room of dimensions 4 $\mathrm{m} \times 7 \mathrm{m} \times 3 \mathrm{m} ?$ (c) Compare your answer in part (b) with the number of atoms you would find in this room under normal conditions on the earth.

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a. 3.5 atoms/m $^{3}$b. 294 atomsc. The post (b) is much less and almost negligible than the number of atoms in a room.

Physics 103

Chapter 30

Nuclear and High-Energy Physics

Atomic Physics

Nuclear Physics

Particle Physics

Cornell University

University of Michigan - Ann Arbor

Simon Fraser University

Lectures

02:42

Atomic physics is the field of physics that studies atoms as an isolated system of electrons and an atomic nucleus. It is primarily concerned with the arrangement of electrons around the nucleus and the processes by which these arrangements change. The theory of quantum mechanics, a set of mathematical rules that describe the behaviour of matter and its interactions, provides a good model for the description of atomic structure and properties.

02:26

In physics, nuclear physics is the field of physics that studies the constituents and interactions of atomic nuclei. The most commonly known applications of nuclear physics are nuclear power generation and nuclear weapons technology, but the research has provided application in many fields, including those in nuclear medicine and magnetic resonance imaging, ion implantation in materials engineering, and radiocarbon dating in geology and archaeology.

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Assuming that the density …

part A of our question says it wants us to assume the universes. All hydrogen expressed the critical density of the number of hydrogen atoms and cubic meters are per cubic meter. So the mass of the hydrogen atom is just gonna be the mass of the proton. It's 1.67 times 10 to the minus 27 kilograms, and the critical density rows of sea is 5.8 times 10 to the minus 27 kilograms per meter cubed. So therefore, to find the number of atoms or the density of the Adams will call this row sub end for the density of the number of atoms. In being the, uh, the number of atoms we find that this here is equal to the mass divided by row sub sea. Plugging those values in, we find that this is equal to 3.473 Uh, and this number here is the number of Adam's per cubic meter. So we can say Adams per meter cubed. We can box this in as the solution apart, eh? Part B says it wants us to find the number in a, uh in a space of a volume of four meters by seven meters by three meters. So the number of atoms that we have we'll call it end is the number density gross of in times the volume V. Well, this is Rosa Been that we found in part A and the volume here is four meters times seven meters times three meters. So carrying out this calculation, we find that the number of atoms that we have here in this room is 294 Adams. We can box it in as their solution to be lastly, Part C says compare your answer that we just found in part B with the number of atoms you'd find in the room under normal conditions on Earth. Okay, so under normal conditions on Earth, we have normal pressure, normal temperature, the number of atoms we'll call this sense of the for the number of atoms in the room on Earth would be about equal to 1.34 times 10 to the 27. This is also the number of atoms. Okay, so therefore, the number of atoms that we calculated is much less than the number of atoms here on earth. All right. And we can box that in as our solution for part C

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