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Homework Statement
Atomic sodium is studied by photoelectron spectroscopy. What is the maximum wavelength of the incident radiation that will ionize the most weakly held electrons in sodium and scatter them so that their de Broglie wavelength measures 3.091e-10 m?
Homework Equations
Energy of photon = h.c/λ
kinetic energy = (1/2)*me*v²
Binding Energy
Ve=h/(me*λe)
The Attempt at a Solution
I've tryed this way but without results
Constants:
h= 6.626068 × 10-34
me = 9.11*10^(-31)
c = 299792458
Ep=Ek+Eb
Ep=1.98762*10^-25*λ^-1
Ek=0.5*9.11*10^(-31)*(6.626068*10^-34/(9.11*10^(-31)*3.091^-10))^2=1.01004*10^-17
The binding energy for the most weakly held electrons i found in this table:
http://xdb.lbl.gov/Section1/Table_1-1.pdf
Eb=30.81 ev = 4,94×10-18 j
thus
1.98762*10^-25*λ^-1=1.01004*10^-17+4,94×10
λ=1.32152*10^-8
Whats wrong?