Max Wavelength of Sodium Ionization

[CFXMSC]

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?

 

[mfb]

The value for the ionization energy is wrong. It should be about 5 eV, not 30 eV.
The table you linked has values for some other electrons, I think.

I did not check your calculations - without any explanation what you are doing, this is not fun.

 

[CFXMSC]

Explanation

But the question is self explanatory. You have a incident photon that will ionize a sodium atom and scatter a electron with a wavelenght equals 3.091e-10. I've just done a energy balance. The energy of the photon must be equal to the kinetic energy of the electron plus binding energy.

Binding energy = 5.1 ev accord with my periodic table from MIT.

 

[CFXMSC]

I solved. I made ​​a mistake in the calculations