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number of electrons in 21Sc with L=1​

Sagot :

jacob193

Answer:

There are [tex]12[/tex] electrons with [tex]l = 1[/tex] in each [tex]\rm _{21}Sc[/tex] atom in its ground state.

Explanation:

The azimuthal quantum number (with symbol [tex]l[/tex]) of an electron in an atom specifies the type of the orbital that contains this electron.

For example:

  • [tex]l = 0[/tex] corresponds to an [tex]s[/tex] orbital.
  • [tex]l = 1[/tex] corresponds to a [tex]p[/tex] orbital.
  • [tex]l = 2[/tex] corresponds to a [tex]d[/tex] orbital.

The question is asking for the number of electrons with [tex]l = 1[/tex] in each [tex]\rm _{21}Sc[/tex] atom. Because [tex]l = 1\![/tex] corresponds to [tex]p[/tex] orbitals, this question is equivalently asking for the number of electrons in [tex]p\![/tex] orbitals in one such atom.

Find the electron configuration of a ground-state [tex]\rm _{21}Sc[/tex] atom:

[tex]1s^{2}\; 2s^{2}\; 2p^{6}\; 3\, s^{2}\; 3\, p^{6}\; 3d^{1}\; 4s^{2}[/tex].

In other words, there are [tex]12[/tex] electrons in [tex]p[/tex] orbitals in total in a ground-state [tex]\rm _{21}Sc[/tex] atom in total; [tex]l = 1[/tex] for each of these electrons because of the

Therefore, a ground-state [tex]\rm _{21}Sc[/tex] atom would include [tex]12[/tex] electrons with [tex]l = 1[/tex].

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