chapter4 - Useful constants, etc.: h := 6.63 10 e := 1.6 10...

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===================================================================================== Coming soon. The answer is 1.46 µ m. What is the shortest wavelength present in the Brackett series of spectral lines? Problem 4.5 Physics 107 ===================================================================================== I will remove this problem from the list of assigned problems. You will not be tested on it. Find the frequency of revolution of the electron in the classical model of the hydrogen atom. In what region of the spectrum are electromagnetic waves of this frequency? Problem 4.4 Physics 107 ===================================================================================== meters r 0 1.137 10 13 × = r 0 Ze 2 4 π ⋅ε 0 K := the eV converts to SI units K1 1 0 6 eV := Z7 9 := Define the constants and variables: r 0 2 4 π 0 K := We can use equation 4.2, modified to account for the +1 charge on the proton. Determine the distance of closest approach of 1 MeV protons incident on gold nuclei. Problem 4.3 Physics 107 ===================================================================================== ε 0 8.85 10 12 := eV 1.6 10 19 := e1 . 6 1 0 19 := c3 1 0 8 := m electron 9.11 10 31 := h6 . 6 3 1 0 34 := Useful constants, etc.: 1
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The desired result. r n n 2 h 2 ⋅ε 0 π m e 2 := r n 2 r n n 2 h 2 0 π m e 2 := r n 2 n 2 h 2 4 π 2 m 2 e 2 4 π 0 m r n := Plug in v from eq. 4.4 r n nh 2 π m v := mv r n := L n hbar := Begin: r n n 2 h 2 0 π m e 2 := and arrive at L n hbar := We want to begin with This problem is not "assigned" for FS 2003. I left the solution here for your interest. You will not be tested on this problem. Lacking de Broglie's hypothesis to guide his thinking, Bohr arrived at his model by postulating that the angular momentum of an orbital electron must be an integral multiple of hbar. Show that this postulate leads to equation 4.13. Problem 4.8 Physics 107 ===================================================================================== The answer to this question is best found in section 4.2. Beiser gives the kinetic and potential energies of the electron in page 1245 The kinetic energy, mv 2 /2, is always positive, and the electron is, indeed, in motion. The potential energy, however, results from the Coulomb attraction of the proton and electron. This negative energy is larger that the kinetic energy. The total energy E=K+V is therefore negative, indicating the electron is bound, but the kinetic energy is positive, so the electron can be in motion. In the Bohr model, the electron is in constant motion. How can such an electron have a negative amount of energy? Problem 4.7 Physics 107 ===================================================================================== Coming soon. The answer is 0.820 µ m. What is the shortest wavelength present in the Paschen series of spectral lines? Problem 4.6 Physics 107 2
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We can calculate the momentum of this electron in several ways. We know its speed, so we can get its momentum from p=mv. In MKS units, a small number. δ p 9.974 10 25 × = δ p h 4 π a 0 () := According to the uncertainty principle, the uncertainty in the momentum of an electron in a ground-state Bohr atom (taking a 0 as an estimate of the size of the region to which the electron is confined) is at least a 0 5.29 10 11 := Compare the uncertainty in the momentum of an electron confined to a region of linear dimension a 0 with the momentum in a ground-state Bohr orbit.
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This note was uploaded on 02/16/2010 for the course CHEM 550 taught by Professor Abra during the Fall '09 term at Pennsylvania State University, University Park.

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chapter4 - Useful constants, etc.: h := 6.63 10 e := 1.6 10...

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