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    pop000's Avatar
    pop000 Posts: 352, Reputation: 6
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    #1

    Mar 16, 2011, 05:03 AM
    Valence electron
    I asked to find the number of Valence electron in Bismuth so my answer is 5. yes?

    And 1 more question is I need to find the number of electrons in 4th level of Au
    So is will be 32? Because if we look on Au in the 4th level so there 2 electrons in 4s 6 electrons in 4p 10 electrons in 4d and 14 electrons in 4f

    Do I correct in my answers ?

    Thanks.
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #2

    Mar 16, 2011, 07:34 AM

    I would say yes :)
    pop000's Avatar
    pop000 Posts: 352, Reputation: 6
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    #3

    Mar 16, 2011, 07:45 AM
    hi.
    thank you for answer.
    so if I asked to find the number of Atomic orbital of: principal quantum number(n) and the orbital Angular momentum quantum number (L) of the orbital 2s


    tell me if I correct n=2 and the Angular momentum L=0 ?

    and in the orbital 6f n=6? And L=3?

    thanks.
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #4

    Mar 16, 2011, 08:00 AM

    Sorry, this time it's really beyond me :(
    pop000's Avatar
    pop000 Posts: 352, Reputation: 6
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    #5

    Mar 16, 2011, 08:47 AM
    Comment on Unknown008's post
    :) that OK still thank you for all your help.
    jcaron2's Avatar
    jcaron2 Posts: 986, Reputation: 204
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    #6

    Mar 17, 2011, 08:53 PM
    hi.
    thank you for answer.
    so if i asked to find the number of Atomic orbital of: principal quantum number(n) and the orbital Angular momentum quantum number (L) of the orbital 2s


    tell me if i correct n=2 and the Angular momentum L=0 ?

    and in the orbital 6f n=6? And L=3?

    thanks.
    Pop, your answers are correct, but be careful. The azimuthal quantum number (also known as the orbital angular momentum quantum number) is usually denoted with a small cursive . Capital L is usually reserved for the angular momentum, itself (or the orbital angular momentum operator in Schrodinger's wave equation), not the quantum number.
    pop000's Avatar
    pop000 Posts: 352, Reputation: 6
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    #7

    Mar 18, 2011, 01:40 AM
    Comment on jcaron2's post
    yes truth I mean L=l.
    thank you. :)

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