Part B How many values of m l are possible for an electron with orbital quantum

Question

Part B

How many values of ml are possible for an electron with orbital quantum number l = 3?

Express your answer as an integer.

The quantum state of a particle can be specified by giving a complete set of quantum numbers (n, 1, mi, ms). How many different quantum states are possible ifthe principal quantum number is n = 2? To find the total number of allowed states, first write down the allowed orbital quantum numbers I, and then write down the number of allowed values of mi 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 Item 9 Learning Goal: To understand and be able to use the rules for determining allowable orbital angular momentum states Several numbers are necessary to describe the states available to an electron in the hydrogen atom. The principal quantum number n determines the energy of the electron. The orbital quantum number l determines the total angular momentum of the electron and the magnetic quantum number mi determines the component of the angular momentum parallel to a specific axis, usually the z axis Submit Hints My Answers Give Up Review Part For a given principal quantum number n, the orbital quantum number can take integer values ranging from zero to n-1. For a given orbital quantum number I, the magnetic quantum number can take integer values from-l to l. A fourth number, the spin mg is important for interactions with magnetic fields and counting states. The spin can be either +1/2 or-1/2, independent of the values of the other quantum numbers. Incorrect; Try Again; 3 attempts remaining Part D Is the state n-3, 1-3. m1--2. m8-1/2 an allowable state? If not, why not? The energy of an electron in hydrogen is related to the principal quantum number by En(-13.60 eV)/n2. The orbital angular momentum is related to the orbital quantum number by L =v/l(l + 1), and the orbital angular momentum in the z direction is related to the magnetic quantum number by O Yes it is an allowable state L-hl+1), and the orbital angular momentum in the z No: The magnetic quantum number cannot be negative O No: The orbital quantum number cannot equal the principal quantum number. No: The magnetic quantum number must equal the principal quantum number. No: The magnetic quantum number must equal the orbital quantum number. Submit My Answers Give Up Part E What is the maximum angular momentum Lax that an electron with principal quantum number n 2 can have? Express your answer in units of h. (You don't need to enter the h, it is in the units field for you.) 8:12 PM

Explanation / Answer

for orbital quantum number l there are 2l+1 possible values of ml

here l=3 there fore 2(3)+1=7 possible ml values

*part c)for n=2 l=n-1=2-1=1,ml=2l+1=2*2+1=5,ms=+1/2

*for l=2 there are 2l+1=2*2+1=5 possible ml states

part d) n=3,l=3,ml=-2,ms=1/2 it is not allowed state bcause ml is not equal to n

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