Calculate the amount of energy not used to produce electricity when a photon of wavelength 1*10^-12m1⋅10−12m hits sodium metal which has a work function of 2.3eV?

1 Answer
Al E.
Jul 15, 2017

E - W_0 = E_DE−W0=ED

Hence, 1.97*10^-13 J1.97⋅10−13J

Explanation:

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)/lambdaE=hcλ
E = ((6.626*10^-34J*m)(2.998*10^8m/s))/(10^-12m)E=(6.626⋅10−34J⋅m)(2.998⋅108ms)10−12m
E = 1.98*10^-13 JE=1.98⋅10−13J

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^-19J2.3eV⋅1.602⋅10−19JeV≈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 J1.98⋅10−13J−3.68⋅10−19J≈1.97⋅10−13J

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

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