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Show the proof of the formula - E n = R H / n² for Hydrogen atom.


2 months ago

ROSHAN MUJEEB
827 Points
							If an electron in an atom is moving on an orbit with periodT, classically the electromagnetic radiation will repeat itself every orbital period. If the coupling to the electromagnetic field is weak, so that the orbit doesn't decay very much in one cycle, the radiation will be emitted in a pattern which repeats every period, so that the Fourier transform will have frequencies which are only multiples of 1/T. This is the classical radiation law: the frequencies emitted are integer multiples of 1/T.In quantum mechanics, this emission must be in quanta of light, of frequencies consisting of integer multiples of 1/T, so that classical mechanics is an approximate description at large quantum numbers. This means that the energy level corresponding to a classical orbit of period 1/Tmust have nearby energy levels which differ in energy byh/T, and they should be equally spaced near that level,{\displaystyle \Delta E_{n}={h \over T(E_{n})}.} [\Delta E_{n}={h \over T(E_{n})}.]Bohr worried whether the energy spacing 1/Tshould be best calculated with the period of the energy state{\displaystyle E_{n}} [E_{n}] , or{\displaystyle E_{n+1}} [E_{n+1}] , or some average—in hindsight, this model is only the leading semiclassical approximation.Bohr considered circular orbits. Classically, these orbits must decay to smaller circles when photons are emitted. The level spacing between circular orbits can be calculated with the correspondence formula. For a Hydrogen atom, the classical orbits have a periodTdetermined byKepler's third lawto scale asr3/2. The energy scales as 1/r, so the level spacing formula amounts to{\displaystyle \Delta E\propto {1 \over r^{3 \over 2}}\propto E^{3 \over 2}.} [\Delta E\propto {1 \over r^{3 \over 2}}\propto E^{3 \over 2}.]It is possible to determine the energy levels by recursively stepping down orbit by orbit, but there is a shortcut.The angular momentumLof the circular orbit scales as√r. The energy in terms of the angular momentum is then{\displaystyle E\propto {1 \over r}\propto {1 \over L^{2}}} [E\propto {1 \over r}\propto {1 \over L^{2}}] .Assuming, with Bohr, that quantized values ofLare equally spaced, the spacing between neighboring energies is{\displaystyle \Delta E\propto {1 \over (L+\hbar )^{2}}-{1 \over L^{2}}\approx -{2\hbar \over L^{3}}\propto -E^{3 \over 2}.} [{\displaystyle \Delta E\propto {1 \over (L+\hbar )^{2}}-{1 \over L^{2}}\approx -{2\hbar \over L^{3}}\propto -E^{3 \over 2}.}]This is as desired for equally spaced angular momenta. If one kept track of the constants, the spacing would beħ, so the angular momentum should be an integer multiple ofħ,{\displaystyle L={nh \over 2\pi }=n\hbar ~.} [{\displaystyle L={nh \over 2\pi }=n\hbar ~.}]This is how Bohr arrived at his model.Substituting the expression for the velocity gives an equation forrin terms ofn:{\displaystyle m_{\text{e}}{\sqrt {\dfrac {k_{\text{e}}Ze^{2}}{m_{\text{e}}r}}}r=n\hbar } [m_{\text{e}}{\sqrt {\dfrac {k_{\text{e}}Ze^{2}}{m_{\text{e}}r}}}r=n\hbar]so that the allowed orbit radius at anynis:{\displaystyle r_{n}={n^{2}\hbar ^{2} \over Zk_{\mathrm {e} }e^{2}m_{\mathrm {e} }}} [r_{n}={n^{2}\hbar ^{2} \over Zk_{\mathrm {e} }e^{2}m_{\mathrm {e} }}]The smallest possible value ofrin the hydrogen atom (Z= 1) is called theBohr radiusand is equal to:{\displaystyle r_{1}={\hbar ^{2} \over k_{\mathrm {e} }e^{2}m_{\mathrm {e} }}\approx 5.29\times 10^{-11}\mathrm {m} } [r_{1}={\hbar ^{2} \over k_{\mathrm {e} }e^{2}m_{\mathrm {e} }}\approx 5.29\times 10^{-11}\mathrm {m}]The energy of then-th level for any atom is determined by the radius and quantum number:{\displaystyle E=-{Zk_{\mathrm {e} }e^{2} \over 2r_{n}}=-{Z^{2}(k_{\mathrm {e} }e^{2})^{2}m_{\mathrm {e} } \over 2\hbar ^{2}n^{2}}\approx {-13.6Z^{2} \over n^{2}}\mathrm {eV} } [E=-{Zk_{\mathrm {e} }e^{2} \over 2r_{n}}=-{Z^{2}(k_{\mathrm {e} }e^{2})^{2}m_{\mathrm {e} } \over 2\hbar ^{2}n^{2}}\approx {-13.6Z^{2} \over n^{2}}\mathrm {eV}]

2 months ago
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