light quanta

The equation explained

ΠE+P′≦Rβν.\Pi E + P' \leqq R \beta \nu .
In wordsThe potential Π times the charge of a gram-equivalent, plus the work per gram-equivalent, is at most the gas constant R times the constant β times the frequency.OverviewDrop the assumption that some quanta give their whole energy to a single electron, and the equation weakens to an inequality: what the electron carries away can only be less than one quantum, never more.Full explanationThe paper says what the preceding equation assumed: that the energy of at least a part of the exciting light's quanta is given up completely, each to a single electron. It calls that assumption a natural one, and here it sets it aside. If an electron may take up only part of a quantum's energy, the fastest electron's energy is no longer fixed at one whole quantum less the work of leaving; it can fall short of it. So the two sides need no longer be equal, and what remains is that they cannot be exceeded. The relation is then one-sided, and the straight line of the preceding paragraphs becomes an upper bound rather than a prediction.Every stepΠE=Rβν−P′\Pi E = R \beta \nu - P'The relation of the preceding display, which assumed that a quantum gives its whole energy to one electron.ΠE+P′=Rβν\Pi E + P' = R \beta \nuMove the work per gram-equivalent to the left. In this form the left side is the whole energy the electron received and the right side is one quantum.ΠE+P′≦Rβν.\Pi E + P' \leqq R \beta \nu .Without the assumption an electron takes up only part of a quantum, so the energy it received is at most one quantum and may be less: the printed line.Also

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