AKG
Sep30-04, 12:16 AM
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?:
E_{electron} = E_{photon} + W
(1) : 2.3 eV = \frac{hc}{200 nm} + W
(2) : 0.90 eV = \frac{hc}{258 nm} + W
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.
Do I have it right?:
E_{electron} = E_{photon} + W
(1) : 2.3 eV = \frac{hc}{200 nm} + W
(2) : 0.90 eV = \frac{hc}{258 nm} + W
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.