Calculate zeff for the 4s electron in a copper atom, cu.
Answers
I'm assuming that you are using Slater's rules for effective nuclear charge. The general idea is to subtract the effect of shielding from the actual nuclear charge, Z. The formula is different for s electrons and d electrons, but the number of protons is the same for the same atom, copper, Z = 29. It is best if you write out the electron configurations out so you can see how many electrons are in each sublevel.
Cu: 1s2 2s2 2p6 3s2 3p6 3d10 4s1
For the 4s electron:
29 - [18(0.85) + 10(1.0)] = 3.7
For the 3d electron:
29 - [9(0.35) + 18(1.0)] = 7.85
You can conclude that the 4s electron experiences a lower effective nuclear charge and is therefore more likely to be lost than the 3d electron, should the copper atom become ionized.
To find: Zeff for the 4s electron in a copper atom, cu.
Solution: Zeff is the effective nuclear charge that is the force that a nucleus apply to the electron
we can write it as
Zeff= Z-S
here Z is the Atomic number of the element
S is shielding constant
For Cu atom
Z will be 29
and configuration will be
1s2 2s2 2p6 3s2 3p6 3d10 4s1
So the electrons with the same group shield will be 0.35
the electrons with n-1 group shield will be 0.85
the electrons with n-2 or lesser group shield will be 1
so, Zeff for 4s electron will be
Zeff= 29 - ( 18(0.85) + 10(1.0) )
Zeff = 3.7
Zeff for the 4s electron in a copper atom, cu will be 3.7.