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GR8677 #19
Problem
 GREPhysics.NET Official Solution Alternate Solutions

Quantum Mechanics$\Rightarrow$}Bohr Theory

Recall the Rydberg energy. QED

Alternate Solutions
 casseverhart132019-10-01 03:22:05 Good Examine and fascinating. Thanks. www.trustytreeservice.comReply to this comment jeka2007-02-17 08:08:36 Energy spectrum of the hydrogen atom is given by the equation $E_n=-\frac{Ry}{n^2}$, where $Ry=13.6\rm{eV}$ is the Rydberg constant. So the right answer is (E)Reply to this comment
casseverhart13
2019-10-01 03:22:05
Good Examine and fascinating. Thanks. www.trustytreeservice.com
ernest21
2019-08-10 03:09:33
Instead of doing the long division by hand, you could just say that gamma is less than 2. Solving the inequality you get that v > \\sqrt{3/4}c. E is the only answer that satisfies this. amino acid game
joshuaprice153
2019-08-08 07:30:34
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jeka
2007-02-17 08:08:36
Energy spectrum of the hydrogen atom is given by the equation

$E_n=-\frac{Ry}{n^2}$,

where $Ry=13.6\rm{eV}$ is the Rydberg constant. So the right answer is (E)
 FortranMan2008-10-16 23:10:09 According to Griffiths, the allowed energies for a hydrogen atom are $E_{n} = -\left[ \frac{m}{2 \hbar^2} \left(\frac{e^2}{4 \pi \epsilon_{0}}\right)^{2}\right] \frac{1}{n^2} = \frac{E_1}{n^2}$ Where $E_{1}$ is the ground state of the hydrogen atom, $E_{1} = 13.6 eV$ The Rydberg constant is defined as $R_{y} = \left[ \frac{m}{4 \pi c \hbar^3}\left(\frac{e^2}{4 \pi \epsilon_{0}}\right)^{2}\right] = 1.097 \times 10^{7} m^{-1}$ Thus the energy levels are given as $E_{n} = - \frac{2 \pi c \hbar R_y}{n^2}$ Not entirely necessary to solve the problem, but it's safer to keep your terms straight.
 VKB2014-03-25 21:44:48 Its a good approach to solve problems @ home,interesting.

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