Amnesia's Dream: Why $\cfrac{1}{3}$ Einstein?

Tuesday, May 5, 2015

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.

Tuesday, May 5, 2015