HOMEWORK #5 ANSWERS
1. Which of the following sets of quantum numbers is allowed for an electron in a one electron
atom?
a. n=4,
l
=3, m=3, m
s
=0
b. n=3,
l
=1, m=2, m
s
= –1/2
c. n=2,
l
=0, m=1, m
s
=1/2
d. n=2,
l
=3, m=3, m
s
=1/2
e. n=6,
l
=5, m= –3, m
s
= –1/2
ANS: E
Lets go through and figure out why a, b, c, and d are incorrect according to the rules we
have for the possible values of the four quantum numbers. (a) we know that m
s
can only
take on values of +1/2 and -1/2, so 0 is not an option. (b) we know that m can only take on
values from –
l
to
l
, so m >
l
is not possible. (c) same reason as b. (d) we know that
l
can
only take on values from 0 to n–
1
, so
l
> n is not possible. The only answer to fit all rules is
(e).
2. Which of the following are possible quantum numbers for a 3d wave function in a hydrogen
atom?
a. n=2
l
=3 m=3
b. n=3
l
=0 m=0
c. n=3
l
=3 m= –2
d. n=3
l
=2 m= –1
e. none of the above
ANS: D
“n” is the principle quantum number (tells you the energy level), which in this problem is 3.
“
l
” is the angular momentum quantum number (tells you the shape of the orbital) and takes
on the integral values from 0 to (n-1) with zero being defined as “s”, and1=p, 2=d, 3=f. So
in this case,
l
= 2 and the answer is d.
3. How many radial and angular nodes does a 4p orbital for a one electron atom have?
a. 3 radial, 1 angular
b. 2 radial, 2 angular
c. 2 radial, 1 angular
d. 2 radial, 0 angular
e. 1 radial, 1 angular
ANS: C
The number of angular nodes is equal to
l
, and the number of radial nodes is equal to n –
l
–
1 (page 182, point #2). Using this information, and that for a p orbital,
l
= 1, the number of
angular nodes must be 1, and the number of radial nodes is 4 – 1 – 1 = 2.