Chem 141 Quiz 6 Fall 2009 Name ...

businessunknownInternet και Εφαρμογές Web

12 Νοε 2013 (πριν από 3 χρόνια και 6 μήνες)

118 εμφανίσεις

Chem 141






Quiz
6







Fall 2009


Name


__________________

________
Key
__
_________
___________ Instructor: Martin Larter


1.

Summarize
the three quantum numbers used to specify atomic orbitals, by using the
symbol
for each,
the possible
values
of each, and the physical
meaning
of each.




n is the principal quantum number, which indicates the energy level. The possible values range from
n=1 to n=





l is the angular momentum quantum number, which determines the shape o
f the orbital. The possible
values range from 0 to n
-
1.




m
l

is the magnetic quantum number, which determines the orientation of the orbital in space. The
possible values range from

l to +l.


2.

For each of the following quantum designations, if it represents possible electrons in an atom write
“correct”. If not, tell where the error take place.


Designation





Possible error


i.

n=1, l=1, m
l
=0, m
s
=
-
1/2



l=1 (should be l=0)



ii.

n=4, l=3, m
l
=
-
4, m
s
=+1/2



m
l
=
-
4 (should be
-
3 to +3)
__



3.

Fill in the blank


a.

Name the rule which states that the most stable configuration of an atom is the one with the
maximum number of electrons with the same spin orientation.

__

Pauli Exclusion Principle

__


b.

Name the
rule which states that if two electrons occupy an orbital, they will have opposite spin

orientations.
___

Hund’s Rule

_____



c.

What does a
node

on an orbital density plot signify?

__

Zero probability of finding an electron

___


4.

Hydrogen has several visibl
e lines in its line spectra. Calculate the wavelength of the green line in the
visible spectrum of excited hydrogen atoms as an electron moves from the n=4 to the n=2 energy level.

(hint:

E
electron

= (

2.18


10

18

J)


[
z
2
/
n
2
])


1/λ = 1.10x10
7
/m(1/n
2
f



1/n
2
i
)



OR

hc/λ =
-
2.18x10
-
18
J(1/n
2
f



1/n
2
i
)



1/λ = 1.10x10
7
/m(.25
-
.0625)





hc/λ = 4.0875x10
-
19
J


1/λ = 2.0625x10
6
/m = 4.8485x10
-
7

m = 485 nm


λ =
6.626x10
-
34

J.s x 2.998x10
8

m/s











4.0875x10
-
19
J









λ = 4.8599x10
-
7

m = 486 nm


















Wavelength

485 nm