Atomic Spectra Relation Cross Scans

Photon Frequency Match Three-Row Scan

Three nearby photon frequencies show that only exact equality fires the line. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

Only the exact frequency closes the photon equality

With the training constant fixed at 1 joule per hertz, frequency 4 hertz gives photon energy 4 joules. The fixed transition gap is 5 joules, so the matched bit is 0.

E=hf=1⋅4=4 J≠5 J,matched=0E=hf=1\cdot 4=4\ \mathrm{J}\ne5\ \mathrm{J},\quad matched=0
Photon match rowNear frequencies stay rejected unless the equality closes.photon=4 Jgap=5 Jmisslowerupper5

Only the exact frequency closes the photon equality

With the training constant fixed at 1 joule per hertz, frequency 5 hertz gives photon energy 5 joules. The fixed transition gap is 5 joules, so the matched bit is 1.

E=hf=1⋅5=5 J=5 J,matched=1E=hf=1\cdot 5=5\ \mathrm{J}=5\ \mathrm{J},\quad matched=1
Photon match rowNear frequencies stay rejected unless the equality closes.photon=5 Jgap=5 Jmatchlowerupper5

Only the exact frequency closes the photon equality

With the training constant fixed at 1 joule per hertz, frequency 6 hertz gives photon energy 6 joules. The fixed transition gap is 5 joules, so the matched bit is 0.

E=hf=1⋅6=6 J≠5 J,matched=0E=hf=1\cdot 6=6\ \mathrm{J}\ne5\ \mathrm{J},\quad matched=0
Photon match rowNear frequencies stay rejected unless the equality closes.photon=6 Jgap=5 Jmisslowerupper5