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]
