It takes 208.4 kJ of energy to remove 1 mole of electrons from the atoms on the surface of rubidium metal. If rubidium metal is irradiated with 254nm light, what is the maximum kinetic energy that a released electron can have? Given: h=6.62x10^(-34) J s c= 3.0x10^8 m/s N=6.02x10^(23) atoms/mol
@matt101 :)
First remember the equation describing the photoelectric effect (which is what you're dealing with here: E = W + KE, where E is the total energy, W (often denoted by ϕ) is the work function of the metal (essentially the minimum energy required to eject an electron from that metal), and KE is the kinetic energy of the electron. The very first sentence of the question gives you the value for W. The second sentence gives you information to find the value for E (remember, the light is supplying ALL the energy, both to free the electrons from the metal and subsequently their kinetic energy for whatever light energy is left). You can find the energy contained in the light using the Planck relation, E=hc/λ. You have E, and you have W, so know you can easily solve for KE. However, keep in mind that the value you find for KE is for 1 mole of electrons, because the W given was for 1 mole of electrons. To find the KE of just 1 electron, you need to divide that value by Avogadro's number. If you have any questions let me know!
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