Which of the following sets of quantum numbers is correct for an electron in 4f orbital?                

  • Option 1)

    n=4,l=3,m=+4,s=+\frac{1}{2}

  • Option 2)

    n=4,l=4,m=-4,s=-\frac{1}{2}

  • Option 3)

    n=4,l=3,m=+1,s=+\frac{1}{2}

  • Option 4)

    n=3,l=2,m=-2,s=+\frac{1}{2}

 

Answers (2)

As we learnt in 

Principal Quantum Number (n) -

 It is a positive integer with value of n = 1,2,3.......

-

 

 

Azimuthal Quantum Number(l) -

 

For a given value of n, l can have n values ranging from 0 to n – 1, that is, for a given value of n, the possible value of l are : l = 0, 1, 2, ....( n –1)

-

Magnetic Quantum Number (m) -

 For any sub-shell (defined by ‘l ’value) 2\: l+1values of m are possible and these values are given by :

m = – l , – ( l –1), – ( l – 2)... 0,1... ( l – 2), ( l –1),l

-

Spin Quantum Number (s) -

It has two values +1/2 and -1/2

-

 

 For 4f\: orbital \: electrons,\: n=4                      

l=3\left ( because\: 0\: 1\: 2\: 3\: \right )

m= +3,+2,+1,0,-1,-2,-3

s= \pm 1/2


Option 1)

n=4,l=3,m=+4,s=+\frac{1}{2}

Incorrect option

Option 2)

n=4,l=4,m=-4,s=-\frac{1}{2}

Incorrect option

Option 3)

n=4,l=3,m=+1,s=+\frac{1}{2}

Correct option

Option 4)

n=3,l=2,m=-2,s=+\frac{1}{2}

Incorrect option

N neha

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