Which would you expect to experience a greater effective nuclear charge? which would you expect to experience a greater effective nuclear charge? a 2p electron of an atom a 3s electron of a na atom?

in atom of sodium (Na) you would expect that a 2p electron experience a greater effective nuclear charge.

Explanation:

 
The effective nuclear charge, Zefff, is the net positive charge experienced by an electron in an atom with more than one electron.

The outer electrons of an atom with several or many electron are simultaneously attracted to the protons in the nucleus and repelled by the negatively charged electrons.

The effective nuclear charge on that electron is calculated using this equation:

Z eff = Z − S, where Z is the number of protons in the nucleus (atomic number), and S is the average number of electrons between the nucleus and the electron in question (the number of non-valence electrons).

For sodium (Na) atomZ is 11, and the electron configuration of the sodium (Na) atom is:

Na (Z=11) : 1s^2 2s^2 2p^6 3s^1

So, the effective nuclear charge of the electrons in the 2p orbitals is greater than the effective nuclear charge of the electron in 3s, given that the there are 10 electrons closer to the nucleus  than the electron in the 3s orbtial, while there are only 4 electrons closer to the nucleus than the 2p electrons.

This is, for a given atom, the further away is one electron from the nucleus the less the effective nuclear charge. That is what is happening with the 3s electron, which is a valence electron.

  remember that hf is only the energy of one photon. not two. not three. one.  1) 5.12 * 10^3 photons  find the energy of one photon, then determine how many photons are needed to reach the sensitivity.  e = hf  = (6.626 * 10^-34 js)(5.95 * 10^19 hz)  = 3.94247 * 10^-14 j per photon  (2.02 * 10^-10 j) / (3.94247 * 10^-14 j/photon) = 5123.69149 photons  round to three sig figs.  2) 5.77 * 10^-17 j  energy of pulse = energy per photon * number of photons  = hfn  = (6.626 * 10^-34 js)(1.66 * 10^13 hz)(5250)  = 5.774559 * 10^-17 j  round to three sig figs.  3) 4.07 * 10^8 kj/mol  10^12 pm = 1 m  c = fλ  f = c/λ  e = hf  = hc/λ  = [(6.626 * 10^-34 js)(3.00 * 10^8 m/s) / (2.94 * 10^-12 m)](1 kj/1000 j)(6.02 * 10^23 photons/mol)  = 40702571.4 kj/mol  round to three sig figs.  4) 9.94 * 10^-1 m  e = hc/λ  λ = hc / e  = (6.626 * 10^-34 js)(3.00 * 10^8 m/s) / (2.00 * 10^-25 j)  = 0.9939 m  round to three sig figs.  5) 9.06 * 10^3 photons  energy of pulse = energy per photon * number of photons  number of photons = energy of pulse / energy per photon  = energy of pulse / (hf)  = (3.87 * 10^-13 j) / [(6.626 * 10^-34 js)(6.45 * 10^16 hz)]  = 9055.23695 photons  round to three sig figs.  6) 2.78 * 10^-15 j  this is problem 1, except in reverse.  sensitivity = minumum number of photons * energy per photon  = (5930)(6.626 * 10^-34 js)(7.077 * 10^14 hz)  = 2.78070758 × 10^-15 j  round to three sig figs.