Quantum Mechanics Question - Work function and h

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SUMMARY

The discussion focuses on calculating Planck's constant (h) and the work function (W) of aluminum using photoelectric effect data. The maximum energy of photoelectrons is given as 2.3 eV for 200 nm radiation and 0.90 eV for 259 nm radiation. The equations E_{electron} = E_{photon} + W are correctly applied, leading to two equations with two unknowns. By subtracting the equations, one can eliminate W to solve for h, and subsequently determine W.

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Quantum Mechanics Question -- Work function and h

The maximum energy of photoelectrons from aluminum is 2.3 eV for radiation of 200 nm and 0.90 eV for radiation of 259 nm. Use these data to calculate Planck's constant and the work function of aluminum.

Do I have it right?:

[tex]E_{electron} = E_{photon} + W[/tex]

(1) : [tex]2.3 eV = \frac{hc}{200 nm} + W[/tex]

(2) : [tex]0.90 eV = \frac{hc}{258 nm} + W[/tex]

Two equations with two unknonws (h and W), so I just solve, right? Seems pretty simple, I'm just not sure if I'm using the right equations.
 
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Some energy of the photon is used to "knock" the electron (work function) and the rest is given to electron as kinetic energy. So
Ephoton = W + KEelectron, so KEelectron = Ephoton - W. Now, having corrected this, we can subtract (1) and (2) part by part, so eliminate W and find h. Then we trivially find W.
 

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