GR 8677927796770177 | # Login | Register

GR8677 #19
Problem
 GREPhysics.NET Official Solution Alternate Solutions

Quantum Mechanics$\Rightarrow$}Bohr Theory

Recall the Rydberg energy. QED

Alternate Solutions
 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
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.

LaTeX syntax supported through dollar sign wrappers $, ex.,$\alpha^2_0$produces $\alpha^2_0$. type this... to get...$\int_0^\infty$$\int_0^\infty$$\partial$$\partial$$\Rightarrow$$\Rightarrow$$\ddot{x},\dot{x}$$\ddot{x},\dot{x}$$\sqrt{z}$$\sqrt{z}$$\langle my \rangle$$\langle my \rangle$$\left( abacadabra \right)_{me}$$\left( abacadabra \right)_{me}$$\vec{E}$$\vec{E}$$\frac{a}{b}\$ $\frac{a}{b}$