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Verbatim question for GR8677 #19
Quantum Mechanics}Bohr Theory

Recall the Rydberg energy. QED

See below for user comments and alternate solutions! See below for user comments and alternate solutions!
Alternate Solutions
casseverhart13
2019-10-01 03:22:05
Good Examine and fascinating. Thanks. www.trustytreeservice.comAlternate Solution - Unverified
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)
Alternate Solution - Unverified
Comments
casseverhart13
2019-10-01 03:22:05
Good Examine and fascinating. Thanks. www.trustytreeservice.comAlternate Solution - Unverified
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 gameNEC
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)
FortranMan
2008-10-16 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.
VKB
2014-03-25 21:44:48
Its a good approach to solve problems @ home,interesting.
Alternate Solution - Unverified

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