Calculate the amount of energy not used to produce electricity when a photon of wavelength #1*10^-12m# hits sodium metal which has a work function of 2.3eV?

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Answer 1 · 2017-07-15T22:44:10

#E - W_0 = E_D#

Hence, #1.97*10^-13 J#

Here we're relating a few things: (1) the energy of the photon, and (2) the difference between the work function and that energy.

#E = (hc)/lambda#
#E = ((6.626*10^-34J*m)(2.998*10^8m/s))/(10^-12m)#
#E = 1.98*10^-13 J#

Above is the energy of the photon, now we're going to convert the work function joules, as such,

#2.3eV*(1.602*10^-19J)/(eV) approx 3.68*10^-19J#

So, relating these two values to realize our differential:

#1.98*10^-13 J - 3.68*10^-19J approx 1.97*10^-13 J#

This is fairly reasonable since a photon of that wavelength is probably gamma, ouch!