Changing the training h rescales photon energy. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

Smaller training h stays below the gap

Frequency stays at 3 hertz while h is 1 J/Hz. The photon energy is 3 joules, so accepted is 0 and excess is 0 joules.

E=hf=13=3 J3 J < Eg=6 Jaccepted=0\begin{array}{c}E=hf=1\cdot 3=3\ \mathrm{J}\\3\ \mathrm{J}\ <\ E_g=6\ \mathrm{J}\\\mathrm{accepted}=0\end{array}
Planck row 1Changing h rescales the checked photon energy.h=1 J/Hzf=3 HzE=3 JE_g=6 Jaccepted=0excess=0 J

Default training h reaches equality

Frequency stays at 3 hertz while h is 2 J/Hz. The photon energy is 6 joules, so accepted is 1 and excess is 0 joules.

E=hf=23=6 J6 J = Eg=6 Jaccepted=1\begin{array}{c}E=hf=2\cdot 3=6\ \mathrm{J}\\6\ \mathrm{J}\ =\ E_g=6\ \mathrm{J}\\\mathrm{accepted}=1\end{array}
Planck row 2Changing h rescales the checked photon energy.h=2 J/Hzf=3 HzE=6 JE_g=6 Jaccepted=1excess=0 J

Larger training h clears the gap

Frequency stays at 3 hertz while h is 3 J/Hz. The photon energy is 9 joules, so accepted is 1 and excess is 3 joules.

E=hf=33=9 J9 J > Eg=6 Jaccepted=1\begin{array}{c}E=hf=3\cdot 3=9\ \mathrm{J}\\9\ \mathrm{J}\ >\ E_g=6\ \mathrm{J}\\\mathrm{accepted}=1\end{array}
Planck row 3Changing h rescales the checked photon energy.h=3 J/Hzf=3 HzE=9 JE_g=6 Jaccepted=1excess=3 J