User:Neil Parker/sandbox

Introduction
As in programming with the "hello world" routine, the first thing to do with a Math rendering system is to display the square root of 2, a 'tradition' established in Ancient Babylon!


 * $$\sqrt{2}$$

Ok - been there, done that, got the t-shirt. Now let's move on to establish my somewhat limited credentials for the following project. Readers will be well advised to take it all with a very big pinch of salt!

Electrical engineer turned Maths/Science tutor - graduated 1985 University of Cape Town. Did Physics I and Physics II. Have a brevity of relativity and a smattering of scattering.

So enjoy the 'crackpot science' but don't say you weren't warned...!! In particular please note that the section on recoil energy was a misguided attempt to explain why the formula did not predict the extremely accurately measured 1s - 2s transition frequency. It was written before I knew anything at all about fine structure and the difference between a 1s - 2s transition as compared to a 1s - 2p 3/2 transition (as we now know, the formula determines the latter). But rather than remove I am leaving that section intact since it is part of the 'evolution' of this model.

Please feel free to click on the "talk" label where there is a freely editable page - I would welcome any comments or suggestions you may have. Actually this page is also "freely editable" (like the rest of Wikipedia!) but can I request that you rather use the talk page for suggestions and let me do the edits here.

Abstract
'
 * We are proposing a model of the Hydrogen atom in which the hydrogen electron is represented by a quantized oscillating energy phasor of the form:


 * $$ E_n = \mu c^2 e^{iw_nt}.e^{i\theta_n}$$ where $$\mu$$ is the reduced mass of an electron occupying the orbital defined by principal quantum number n.


 * $$w_n=\frac{2\pi\alpha c}{n\lambda_e}$$ with $$\lambda_e$$ in this context taken as the Compton wavelength of a 'reduced mass' electron bound to the proton nucleus of the hydrogen atom and $$\alpha$$ is the fine structure constant.


 * $$\theta_n=\cos^{-1}\left(\frac{\alpha}{n}\right)$$ and average 'wave energy' of quantum state n is $$\Im \left( \mu c^2 e^{i\theta_n}\right)=\mu c^2sin(\theta_n)=\mu c^2\sqrt{1-\frac{\alpha^2}{n^2}} $$.

.
 * This means that photons (ie wave energy) emitted or absorbed during a transition from energy level n to energy level m (n>m) have frequency given by the equation:


 * $$h\Delta f =h(f_n-f_m)= \mu c^2\left(sin(\theta_n)-sin(\theta_m)\right)=\mu c^2\left[\sqrt{1-\frac{\alpha^2}{n^2}}-\sqrt{1-\frac{\alpha^2}{m^2}}\right]$$


 * where h is the Planck constant.


 * We show that the above formula corresponds to a subset of the Dirac energy levels for an electron bound to the Hydrogen nucleus.


 * A table of values showing calculated transition frequencies (as compared to values obtained from NIST) can be viewed here. The +- 8.15GHz differences in the Lyman series values are largely attributable to the so-called "Ground State Lamb Shift" - about 8.173 GHz. For greater accuracy, one should also employ a somewhat more complex second order equation as per theory of reduced mass. However even as things stand the differences are measured in parts per million for the Lyman series values and in parts per billion for the Balmer series values.

Newton's Second Law
Force = Change of Momentum with Change of Time (I am taking the definition stated on the | Nasa Website). Mathematically the Second Law takes on various forms depending on circumstance:


 * $$ F=\frac{m_1v_1-m_2v_2}{t_1-t_2} $$
 * $$ F=m\frac{dv}{dt}=ma $$ (this assumes constant mass)
 * $$ F=\frac{d(mv)}{dt}=\frac{dp}{dt} $$ (p represents momentum)

One should bear in mind that force, velocity, acceleration and momentum are all vector quantities so technically all the above are vector equations. In the discussion which follows we will try to keep things simple by assuming all the vectors are in the same direction.

Relativistic Mass and Newton's Second Law
In relativity mass varies with velocity according to the equation:


 * $$m = {m_0\over \sqrt{1-\displaystyle{v^2\over c^2}}}.$$

So we pose the question what if - in the second of the Newton II equations above - we replace mass with relativistic mass ? Plainly we are no longer assuming 'constant mass'.


 * $$F = {m_0\over \sqrt{1-\displaystyle{v(t)^2\over c^2}}}\frac{dv(t)}{dt} $$

Looks somewhat complex so let's try feeding the equation to a 'math engine' namely Wolfram Alpha.

The somewhat remarkable result is that the above has a mathematically straightforward solution for v(t) in the form of a wave equation!

$$ v(t) = c.sin\left (\frac{Ft+k_1m_0}{m_0c} \right)$$

Now we wonder does this somewhat bizarre result make any physical sense ? What it seems to be saying is that if a mass is accelerated very rapidly towards the speed of light, it will experience some form of wave motion. But isn't that exactly what Modern Physics is telling us - namely that associated with an object having momentum p, we have the | De Broglie wavelength:


 * $$\lambda = \frac{h}{p}.$$

So perhaps we should not be surprised to find that a very rapidly accelerated particle behaves as if it were a wave.

Expansion of the wave equation
Of some interest is the Taylor Series expansion of the v(t) wave equation. Once again we call upon "the Oracle" namely Wolfram Alpha to effect | the expansion (leaving out the constant of integration) and are perhaps somewhat 'underwhelmed' to find that - discarding third order and higher terms - we obtain:

$$v(t) \approx \frac{Ft}{m_0} \approx at $$ as per conventional equations of motion.

Nonetheless it is encouraging to note that the small angle approximation of the wave equation has a familiar form.

Note that the frequency of the wave equation has c (velocity of light) in the denominator and therefore it is clear that wave behavior will only manifest if the F/m ratio is extremely high.

$$ f= \frac{\omega}{2 \pi}= \frac{ F}{2 \pi m_0 c} $$

For example an object falling under gravity will most certainly not show any signs of oscillatory motion unless perhaps falling in the vicinity of a black hole!

Integration of the wave equation
An obvious next step is to integrate the v(t) wave equation to obtain an expression for displacement x(t):

$$x(t)= \int c.sin\left ( \frac{Ft}{m_0c} \right )dt=\frac {-m_0c^2}{F}cos\left ( \frac{Ft}{m_0c} \right ) + constant$$

It's clear that mathematically we had expected another wave equation for x(t) but how are we to interpret an equation which presents us with some form of 'bounded motion' ? Noting that mc^2 appears in the numerator and F in the denominator, it seems sensible to multiply through by the fixed force F to obtain instead a 'bounded energy' equation. The maximum value thereof is none other than Einstein's iconic E = mc2.

$$ F.x(t) = -m_0c^2cos \left (\frac{Ft}{m_0c} \right )$$

Once again it is useful to obtain a small angle approximation for the above energy expression:

$$ F.x(t) \approx -m_0c^2 + \frac{m_0}{2}\left ( \frac{Ft}{m_0} \right )^2 \approx -m_0c^2+ \frac{m_0.v(t)^2}{2} $$

We are using the previously obtained small angle approximation for v(t) as Ft/m. It is not clear quite how we should interpret the signs of the two terms in the above expression but it seems that mc2 is the maximum energy of a particle having mass m from which kinetic energy ½mv(t)2 is deducted since the signs of the respective terms are opposite. But once again we are finding a familiar quantity - namely kinetic energy - emerging from the wave equation.

Energy however cannot 'oscillate' - it must rather change from one form to another just as in a pendulum there is a constant transfer of energy between potential and kinetic form. So we must presume that corresponding to the oscillatory cosine above, there is an oscillatory sine such that:

$$ m_0c^2 = m_0c^2\left[cos^2 \left (\frac{Ft}{m_0c} \right )+sin^2 \left (\frac{Ft}{m_0c} \right )\right]^{1/2}$$

We conjecture that the cosine expression above represents 'conventional' or 'particle' energy whilst the sine is radiant or wave energy. And that the 'oscillation' represents a continual changing from one form to the other.

Complex Energy Equation
Alternatively we could propose a complex energy equation having the following form:

$$E=m_0c^2\exp{\left(\frac{iFt}{m_0c}\right)}$$

The interesting aspect of this equation is its Taylor Series expansion which we can obtain from Wolfram Alpha by entering "expand m*c^2*exp(i*(F*t)/(m*c)" on the command line. The result is displayed below:



Discarding third order and higher terms and noting that the product "Ft" has dimensions of momentum (p) we have a term $$mc^2$$ representing rest mass energy followed by an imaginary term of the form $$pc$$ (wave energy ?) and then a term of the form $$\frac{p^2}{2m} $$ (kinetic energy). The above equation stemmed from this discussion on Physics forums.

The Hydrogen Atom: Bohr Model
We next search for a possible physical situation where the wave equation could apply. A 'candidate mass' would most likely be extremely small and subject to a relatively massive constant force such that the ratio F/m is large. We consider one such situation - an electron in Hydrogen's 1s orbital. One needs to have some appreciation in this instance of just how massive the electrostatic force (between electron and nucleus) is relative to electron mass so we begin with a brief summary of the Bohr model.

The first equation Bohr drew up is based on the electrostatic force on an 'orbiting' electron being equal to the centripetal force:

$$ \frac{m_ev^2}{r}=\frac {kq^2}{r^2} $$

He next assumed that an 'orbiting' electron with De Broglie wavelength h/p would form a standing wave. Thus:

$$ 2 \pi r = n \lambda = \frac {nh}{m_ev} $$

We request Wolfram Alpha to | solve simultaneously for r and v obtaining:

$$r=a_0=\frac{h^2n^2}{4\pi^2m_e k q^2}=5.29177\times 10^{-11}m\text{ (meters)}$$

Introducing the symbol $$\bar{\lambda}_e$$ for the | reduced Compton wavelength (for the electron), the Bohr radius $$a_0$$ may be written as $$\frac {\bar{\lambda}_e}{\alpha} $$

where α is the fine structure constant.

$$v=\frac {2\pi k q^2}{hn}=\alpha c = 2.187691\times10^{6} m/s \text{ (meters per second)} $$

The Bohr model formulae have of course been superceded by the Schrodinger wave equation but we will use the above to provide some idea of the F/m ratio for the electron in the Hydrogen 1s orbital:

$$\frac{F}{m_e}=\frac {kq^2}{r^2m_e}=9.04422\times10^{22} m/s^2 \text {(meters per second squared)} $$

Verify this calculation here.

It is plain that an acceleration of this magnitude is going to result in the electron exhibiting the oscillatory behavior predicted by the v(t) wave equation referred to above. Indeed we can calculate the frequency thereof:

$$f=\frac {k q^2}{2 \pi r^2 m_e c}=4.80143\times 10^{13} Hz \ (hertz)$$

Again verify this calculation here

The Hydrogen Atom: Oscillatory Model
As in the Bohr model, our first equation assumes the normal equation for an orbiting mass:
 * $$ \frac{m_e v^2}{r}=\frac {kq^2}{r^2} $$

But for the second equation - in keeping with the predicted oscillatory behaviour of the electron - we will deploy the standard wave equation v = fλ with the wavelength being the electron's De Broglie wavelength h/mv and the frequency being that predicted by the v(t) wave equation determined previously:


 * $$v=f \lambda = \frac{kq^2}{2 \pi r^2 m_e c} \times \frac{h}{m_e v} $$

Probably one has to invoke 'crackpot science' to justify the marriage of a "particle" equation (mv2/r = kq2/r2) with a wave equation (v = fλ). But after all isn't that precisely what Bohr himself did ?!

As was done for the Bohr model, we submit these two equations to Wolfram Alpha to solve simultaneously for r and v obtaining:


 * $$r=\frac{h}{2 \pi m_e c}=3.861593\times 10^{-13} \text{ m (meters)} $$

and
 * $$v=\sqrt{\frac{2 \pi k q^2}{h c}} \times c=\sqrt{\alpha}c=2.56096338 \times 10^{7} \text{ m/s (meters per second)} $$

The very interesting result which arises is that r turns out to be the reduced Compton wavelength for an electron. The 'electron wave' thus occupies an orbital circumference of length equal to the Compton wavelength h/mc. v is also determined but in this context is somewhat difficult to interpret since an 'electron wave' occupying the Compton wavelength must be travelling at c, the speed of light, in order to form a standing wave.

'Vis Viva' and Wave Energy of the Hydrogen 1s electron
Returning to the equation:


 * $$ \frac{m_e v^2}{r}=\frac {kq^2}{r^2} $$

We note that if we multiply through by r on both sides we obtain:


 * $$ {m_e v^2} = \frac {kq^2}{r} $$

The right hand side of this equation is the electron's potential energy and we should therefore consider the quantity mv2 (the 'vis viva') on the left hand side as the electron's 'conventional' or 'particle' energy. Picking up our earlier determination of oscillating energy with a bounded maximum E=mec2 we surmise that this energy is in quadrature with the electron's wave energy - that is it satisfies the following Pythagorean relationship:


 * $$ \left(m_ev^2\right)^2+\left(hf\right)^2 = (m_e c^2)^2 $$



The above equation may be re-arranged to obtain an expression for wave energy.


 * $$ hf = \sqrt{\left (m_e c^2\right)^2 - \left(m_ev^2\right)^2} $$

Substituting v2 = αc2 - as determined above - we obtain:


 * $$ hf = m_e c^2 \sqrt{1 - \alpha^2} $$

Hydrogen 1s-2s and 1s-3s Transition Energies
In the oscillatory model outlined above, quantization of orbitals arises from the requirement for the orbital circumference(s) to accomodate standing waves based on the electron's Compton wavelength h/mec. Thus allowed radii satisfy the simple condition:

$$r_n=n\bar{\lambda}_e$$

As will be recalled from an earlier section $$\bar{\lambda}_e$$ is the reduced Compton wavelength. Considering the case where n=2, the radius has doubled and therefore the electron's potential energy or 'vis viva' has halved. The expression for wave energy in the 2s orbital is thus:


 * $$ hf_2 = m_e c^2 \sqrt{1 - \left(\frac{\alpha}{2}\right)^2}= m_e c^2 \sqrt{1 - \frac{\alpha^2}{4}} $$

and in general:


 * $$ hf_n = m_e c^2 \sqrt{1 - \left(\frac{\alpha}{n}\right)^2}= m_e c^2 \sqrt{1 - \frac{\alpha^2}{n^2}} $$

We are now in a position to determine the difference in wave energy (transition energy) between the Hydrogen 1s and 2s orbitals:


 * $$ hf_2-hf_1 = m_e c^2 \left[\sqrt{1 - \frac{\alpha^2}{4}}-\sqrt{1-\alpha^2}\right] $$

This is a calculation we shall entrust once more to Wolfram Alpha. Note that in the calculation we use the electron's "reduced mass" as is required for two body problems (in this case proton/electron). Using the standard equation for wave energy (E=hf), we determine the frequency corresponding to this transition energy. The result 2.46608×1015 Hz (hertz) differs from the accurately measured value


 * $$f_{1S-2S} = \text{2 466 061 413 187 035 (10) Hz}\,$$

by about 8 parts per million.

Similarly we may determine the transition energy betweeen the Hydrogen 1s and 3s orbitals:


 * $$ hf_3-hf_1 = m_e c^2 \left[\sqrt{1 - \frac{\alpha^2}{9}}-\sqrt{1-\alpha^2}\right] $$

Again we convert to frequency using E=hf. The result f1s-3s = 2.92276×1015 Hz (hertz) differs from the accurately measured value


 * $$f_{1S-3S} = \text{2 922 742 936.729 (13) MHz}\,$$.

by about 4 parts per million.

Recoil Energy Correction
Conventionally the physics of transition energies in the Hydrogen atom does not take into account recoil energy since it is assumed that recoil occurs against the very large combined mass of nucleus and electron. In this case any theoretical recoil is vanishingly small. However in this model, we note that the wave energies obtained for the 1s-2s and 1s-3s transitions are marginally larger than those corresponding to the accurately measured frequencies for these transitions. We therefore conjecture that it may be possible for the transiting electron to absorb recoil energy independent of the nucleus - the idea is that the fixed transition energy 'budget' is the sum of emitted wave energy and electron recoil energy.

Noting that photon emission is an entirely random process in respect of direction, we further conjecture that only recoil from emissions that are in the plane of the electron's orbital need be considered - any photon emission perpendicular to this (ie along a radial line centred at the nucleus) will indeed recoil against the combined mass of electron and nucleus and thus may be considered recoil free emission.

With this in mind - using conservation of momentum with θ being the random angle (measured from the orbital plane) at which photon emission occurs - we write:

$$\Delta\rho_e=\rho_p=\frac{hf}{c}cos(\theta)$$

$$\Delta v_e = \frac{\Delta\rho_e}{m_e}=\frac{hfcos(\theta)}{m_e c}$$

$$\Delta E = \frac{1}{2} m_e\Delta v_e^2 = \frac{1}{2}m_e \left({\frac{hfcos(\theta)}{m_e c}}\right)^2=\frac{cos^2(\theta)}{2 m_e c^2} (hf)^2 $$

Transition energy Et is the sum of emitted wave energy plus recoil energy - thus:

$$ E_t=hf+\frac{cos^2(\theta)}{2 m_e c^2} (hf)^2 $$

Solving the quadratic in hf and using a two term binomial expansion of the resultant surd, we obtain:

$$ hf = E_t - (E_t)^2 \left( \frac{cos^2(\theta)}{2 m_e c^2}\right) $$

Finally the average value of cos2(θ) over the interval 0 to π is ½ and the recoil corrected equation reduces to:

$$ hf=E_t-\frac{(E_t)^2}{4 m_e c^2} $$

Applying the recoil correction, we obtain f1s-2s=2.46607x1015 Hz and f1s-3s=2.92274x1015 Hz. Both results now differ from the accurately measured values by less than 3 parts per million. The recoil energy corrections have proven to be of the exact order of magnitude required to further reduce the previously determined error margins.

It is appreciated that photon emission is no doubt a somewhat more complicated stochastic process than this simple model implies and may well differ depending on the transition in question. So the above derivation is once again the best that 'crackpot science' can offer.

Deuterium 1s-2s Transition
Deuterium is an isotope of Hydrogen with an atomic mass which is approximately double that of the latter. A small frequency difference between the 1s-2s transitions of hydrogen and deuterium is expected due to the difference in the reduced mass of the electron as is deployed in all the formulae discussed above. In the same article which describes the accurate measurement of the Hydrogen 1s-2s transition at 2 466 061 413 187 035 (10) Hz, CG Parthey also measures the frequency difference between Hydrogen and Deuterium which he gives as 670 994 334 606(15) Hz. Using the Codata value for the atomic mass of Deuterium, our formula developed above yields a frequency difference of 671.012 GHz which is in good agreement (+ 18 MHz) when compared to Parthey's measurement.

For spectroscopic applications it is useful to convert this frequency difference into a wavelength difference:

$$\, \Delta f = f_D-f_H $$

$$ \Delta \lambda = \frac {c}{f_D}-\frac{c}{f_H}=\frac{c}{f_H+\Delta f}-\frac {c}{f_H}=\frac{c(1+\frac{\Delta f}{f_h})^{-1}}{f_H}-\frac {c}{f_H} $$

Using a one term binomial expansion of the expression in brackets, we obtain:

$$\Delta \lambda=\frac{-c \Delta f}{{f_H}^2} $$

Using Parthey's measured values for the frequency difference and H 1s-2s transition frequency we obtain a wavelength difference of 33.08 picometers.

Measuring Lamb Shift: | The Herzberg Experiment
In deference to Dr Herzberg - an outstandingly meticulous experimental physicist - we shall simply present his own summary of an experiment carried out in the mid fifties. This experiment was designed to obtain a highly accurate value of the 1s - 2p transition wavelength for Deuterium (in absorption) so that a measured value of the theoretical Lamb Shift could be obtained.


 * According to modern quantum electrodynamics, the 1S level (ground state) of hydrogen and deuterium is predicted to lie 0,272 cm-l above the energy given by the Dirac theory. In order to obtain an experimental value for this Lamb shift, the absolute wavelength of the L&alpha; line of deuterium has been determined by means of a 3 m vacuum grating spectrograph in fifth order. Lines of the series 61S0-nlPl and 61S0-n3Pl of 198Hg observed in the same order were used as standards. The wavelengths of these standards were obtained to +0.0002A by the combination principle from lines in the visible and near ultra-violet regions, some of which were newly measured. Both the far ultra-violet Hg lines and La of deuterium were simultaneously recorded on the plate in absorption. In this way, from six independent plates the wavelength of the L&alpha; line of deuterium was found to be 1215.3378 +_ 0.0003 A. This value refers to an unresolved doublet. If the relative intensity of the two doublet components (2:1)and the temperature of the absorbing column (80°K) is taken into account and the result compared with the Dirac value of L&alpha;, a shift of the l 2S level of 0.26 +0.03cm-l is obtained. The agreement with the predicted value is very satisfactory.

Using the theory developed above we compare our theoretical value for L&alpha; against the Dirac value as obtained in wave number form in the above experiment. We also deduct observed value from theoretical value to determine a value for Lamb Shift based upon our formula. The Dirac wave number for the Deuterium L&alpha; transition as per Herzberg's report is 82281.933 cm-l whilst that predicted by our formula is 82281.939 cm-l. Deducting Herzberg's very carefully obtained experimental value of 82281.67 cm-l determines values of 0,263 cm-l and 0,269 cm-l respectively for measured Lamb shift. These both compare favourably against the predicted Lamb Shift value of 0.272 cm-l as per Herzberg's summary above.

Measuring the Balmer Alpha Line: | Laser Saturation Spectroscopy
In the seventies a group of physicists (T. W. Hansch, M. H. Nayfeh, S.A. Lee, S.M. Curry,  and I.S. Shahin) at Stanford University set out to establish a considerably more accurate value of the Rydberg constant by measuring - to a very high degree of precision - the Balmer Alpha lines of both Hydrogen and Deuterium. Once again we will quote verbatim from these scientists' report in Physical Review Letters Volume 32, Number 24 17 June 1974:

''We have determined the Rydberg constant with an almost tenfold improvement in accuracy over other recent experiments, by measuring the absolute wavelength of the optically resolved 2P3/2-3D5/2 component of the red Balmer line H&alpha; of atomic hydrogen and D&alpha; of deuterium in a Wood discharge. Doppler broadening was eliminated by saturation spectroscopy' with a pulsed, tunable, dye laser. An iodine-stabilized He-Ne laser served as the wavelength standard. The same measurements provide a new precise value for the H-D isotope shift. We have also determined the splittings between the stronger fine structure components in the optical spectrum to within a few megahertz.''

The wave numbers obtained (by the Stanford group) for the H&alpha; and D&alpha; (Balmer series) lines were 15233.0721(9) cm-1 and 15237.21538(8) cm-1 respectively. The formula we have developed yields predicted values of 15233.07094 cm-1 and 15237.21585 cm-1 respectively and it will be seen that the deviation from measured occurs only in the eighth significant figure for Hydrogen and the ninth for Deuterium. We suspect that even these small discrepancies may be explained by the Lamb shift value(s) pertinent to the 2P3/2-3D5/2 transition.

Subtracting the formula values for Hydrogen and Deuterium (Balmer Alpha lines), we obtain a value of 4.14491cm-1 for the H-D isotope shift. This is in good agreement with the reported measured value of 4.14517cm-1.

Calculation of the Balmer Alpha lines for Hydrogen and Deuterium is shown in this spreadsheet.

Dirac Energy Simplification
Since the above formula yielded results very closely comparable to accurately measured values, one was led to suspect that it must be very similar to the Dirac Energy formula:


 * $$\begin{array}{rl} E_{n\,j} & = \mu c^2\left(1+\left[\dfrac{Z\alpha}{n-|k|+\sqrt{k^2-Z^2\alpha^2}}\right]^2\right)^{-1/2}\\ &\\& \end{array}$$

Following queries submitted to the online Physics Forums and guidance thereby obtained (see this link), the following was determined:

For the particular case(s) where for any given principal quantum number n, we set orbital angular momentum quantum number $$l_{max} = n-1$$ and in turn set total angular momentum quantum number $$j_{max}=l+\frac{1}{2}$$, then $$k = j+\frac{1}{2} = l+1 = n $$. Thus n - |k| = 0 and the expression simplifies:


 * $$\begin{array}{rl} E_{n\,j} & = \mu c^2\left(1+\left[\dfrac{Z\alpha}{\sqrt{n^2-Z^2\alpha^2}}\right]^2\right)^{-1/2}\\ &\\& \end{array}$$
 * $$E_{n\,j} = \mu c^2\left(\frac{n^2}{n^2-Z^2\alpha^2}\right)^{-1/2}=\mu c^2\sqrt{1-\frac{Z^2\alpha^2}{n^2}} $$

So for this particular circumstance, Dirac energy is exactly equivalent to the formula determined above. It is presumed we are dealing with Hydrogen and therefore that Z=1 in the formula. The author would like it to be clear that he has zero understanding of how the Dirac energy is conventionally obtained nor indeed of "orbital angular momentum quantum number". The simplification obtained above is simply a result of suspecting a similarity between equations that give the same (or closely similar) numeric results and then noting that in certain circumstances n -|k| = 0 whereafter the Dirac energy expression can be considerably simplified using straightforward algebra.

A consequence of the above correspondence is that the formula we have developed is 'valid' for the Deuterium Lyman alpha transition studied by Herzberg ( a 1s - 2p 3/2 transition) and the Hydrogen Balmer alpha transition (2p 3/2 - 3d 5/2) obtained by laser absorption spectroscopy. But not for the very accurately determined 1s - 2s transition unless we 'borrow' an adjustment for the 2s - 2p 3/2 difference. In general the formula is accurate for transitions from the highest energy level of one principal quantum number to the highest of another.