Question

Part C

The quantum state of a particle can be specified by giving a complete set of quantum numbers (n,l, ml,ms). How many different quantum states are possible if the principal quantum number is n = 2?

To find the total number of allowed states, first write down the allowed orbital quantum numbers l, and then write down the number of allowed values of ml for each orbital quantum number. Sum these quantities, and then multiply by 2 to account for the two possible orientations of spin.

Express your answer as an integer.

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Part D

Is the state n=3, l=3, ml=?2, ms=1/2 an allowable state? If not, why not?

Is the state , , ,  an allowable state? If not, why not?

A) Yes it is an allowable state.

B) No: The magnetic quantum number must equal the orbital quantum number.

C) No: The magnetic quantum number must equal the principal quantum number.

D) No: The orbital quantum number cannot equal the principal quantum number.

E) No: The magnetic quantum number cannot be negative.

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Part E

What is the maximum angular momentum Lmax that an electron with principal quantum number n = 5 can have?

Express your answer in units of ?. (You don't need to enter the ?, it is in the units field for you.)

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Lmax =

  ?  

Answer 1

Part C Answer:

For Principal quantum number n=2, orbital quantum number l=1.For l=1, m_l can have (2l+1)=(2x1+1)=3 values and they are (-1,0,+1). For each m_l  there are 2 values of m_s(spin quantum number)=(+-1/2) making 6 different combinations of (n,l,m_l,m_s). These are:

(2 ,1,-1,+1/2), (2,1,-1,+1/2), (2,1,0,+1/2),(2,1,0,-1/2),(2,1,1,+1/2),(2,1,1,-1/2).

Part D Answer : Option (D)No: The orbital quantum number cannot equal the principal quantum number.

For n=3 ,the value of l can be 0 l n-1 i.e (0,1,2) so l 3.

Part E Answer:

For n= 5 , l= 4,

L_max = sqrt(l*(l+1))h/2pi = sqrt(4*(4+1))h/2pi =1.42 h.