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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 ml determines the component of the angular momentum parallel to a specific axis, usually the z axis.

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 l, the magnetic quantum number can take integer values from −l to l. A fourth number, the spin ms, 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.

The energy of an electron in hydrogen is related to the principal quantum number by En=(−13.60eV)/n^2. The orbital angular momentum is related to the orbital quantum number by L=ℏ√l(l+1), and the orbital angular momentum in the z direction is related to the magnetic quantum number by Lz=mlℏ.

Required:
a. How many different values of I are possible for an electron with principal quantum number n= 5?
b. How many values of mi are possible for an electron with orbital quantum number I= 3?
c. The quantum state of a particle can be specified by giving a complete set of quantum numbers (n, l, mi, ms). How many different quantum states are possible if the principal quantum number is n= 3?
d. Is the state n =3, I =3, m= —2, ms =1/2 an allowable state? If not, why not?
e. What is the maximum angular momentum Lmax that an electron with principal quantum number n= 3 can have?

Respuesta :

Answer:

a) The number of values of l = 5

b) Number of values of [tex]m_{l}[/tex] = 7

c) Total number of quantum states possible = 18

d) The state is not an allowable state, maximum value of l = n -1

e) [tex]L_{max} = \sqrt{6 }h'[/tex]

Explanation:

a) If the principal quantum number = n

The orbital quantum number, l will range from 0 to n-1

For principal quantum number, n = 5, l will range from 0 to 4 (i.e. 0,1,2,3,4)

The number of values of l = 5

b)for an electron with orbital quantum number, l

The values of [tex]m_{l}[/tex] will range from -l to +l

Therefore, for l = 3, [tex]m_{l}[/tex] = -3, -2, -1, 0, 1, 2, 3

it can be from -3 -2 -1 0 1 2 3  

Number of values of = 7

c) Number of quantum states possible for a principal quantum number, n = 3

The subsidiary quantum number, l = 0 , 1 , 2

When l = 0, m = 0

When l = 1, m = -1, 0, 1

When l =2, m = -2,-1,0,1,2

There is 1 state for l =0, 3 states for l = 1, 5 states for l = 5

The total number of quantum states possible = 2( 1 + 3 + 5+) = 2 * 9

Total number of quantum states possible = 18

4) orbital quantum number always takes values from 0 to n-1, orbital quantum number cannot have the same vale with the principal quantum number, It only has the highest value of n -1, the state described is not allowable.

5) n = 3

For n = 3, maximum value of the orbital quantum number is l = 3 -1

l = 2

The maximum angular momentum is given by the relation:

[tex]L_{max} = \sqrt{l(l+1)} h' \\L_{max} = \sqrt{2(2+1) } h'\\L_{max} = \sqrt{6 }h'[/tex]