Why Does the Lone Electron in Silver Have Zero Orbital Angular Momentum?

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"There are 47 electrons surrounding the silver atom nucleus, of which 46 form a closed inner core of total angular momentum zero – there is no orbital angular momentum, and the electrons with opposite spins pair off, so the total angular momentum is zero, and hence there is no magnetic moment due to the core. The one remaining electron also has zero orbital angular momentum"
Can anyone tell me why the remaining electron has zero orbital angular momentum?
Thanks!
 
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It's in the ground state, so that it has zero orbital angular momentum.
 
ENDLESSYOU said:
"There are 47 electrons surrounding the silver atom nucleus, of which 46 form a closed inner core of total angular momentum zero – there is no orbital angular momentum, and the electrons with opposite spins pair off, so the total angular momentum is zero, and hence there is no magnetic moment due to the core. The one remaining electron also has zero orbital angular momentum"
Can anyone tell me why the remaining electron has zero orbital angular momentum?
Thanks!

If you google "silver electron configuration", you'll get the answer that it is Kr 4d10 5s1. So, "Kr" tells you that silver electron configuration is like krypton's plus additional 11 electrons. Ten of them fill up 4d shell completely which does not have any angular momentum because there are 5 electrons in each spin orientation and sum of the orbital angular momentum of all electrons in d shell cancels out exactly. All that is left is 11th electron which does not have orbital electron because it is in s shell, so total angular momentum of silver atom is equal to spin of the 5s1 electron.
 
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