In a similar way, in photoelectric effect, only \(p_{t_T}\)(\(x_1\),\(x_2\),\(t_c\)) and \(p_{t_g}\)(\(x_1\),\(x_2\),\(t_c\)) will eject electrons from the metal by imparting electric potential energy upon them.
The remaining \(\cfrac{2}{3}\) (\(p_{t_c}\)(\(x_1\),\(x_2\),\(t_g\)), \(p_{t_c}\)(\(x_1\),\(x_2\),\(t_T\)), \(p_{t_T}\)(\(x_1\),\(x_2\),\(t_g\)) and \(p_{t_g}\)(\(x_1\),\(x_2\),\(t_T\)) ) will not eject electrons. \(p_{t_c}\)(\(x_1\),\(x_2\),\(t_g\))/\(p_{t_c}\)(\(x_1\),\(x_2\),\(t_T\)) is the partial charge measured/calculated previously.
In this way we account for the missing \(\cfrac{2}{3}\) of the total photon count illuminating the metal in the photoelectric experiment.
Hello Albert again.
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