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The possible quantum number for $3d$ electron are
A
$n = 3,\,l = 1,\,{m_l} = + 1,\,{m_s} = - \frac{1}{2}$
B
$n = 3,\,l = 2,\,{m_l} = + 2,\,{m_s} = - \frac{1}{2}$
C
$n = 3,\,l = 1,\,{m_l} = - 1,\,{m_s} = + \frac{1}{2}$
D
$n = 3,\,l = 0,\,{m_l} = + 1,\,{m_s} = - \frac{1}{2}$
Solution
For an electron in $3 d$ orbital, the possible values of the quantum numbers are $n =3,1=2, ml =-2,-1,0,+1,+2$
Standard 12
Physics
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