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1 Question 1
Calculate the energy carried by a photon of frequency 1
.
0
×
10
15
Hz
(
h
= 6
.
6
×
10

34
J
·
s
).
E
photon
=
hf
= 6
.
6
×
10

34
J
·
s
×
1
.
0
×
10
15
Hz
= 6
.
6
×
10

19
J
2 Question 2
What is the momentum of an electron whose energy is 1000eV? What is the
de Broglie wavelength? (1
eV
= 1
.
6
×
10

19
J
) Mass of the electron,
m
e

=
9
×
10

31
kg
.
E
e
=
p
2
2
m
e
→
p
e
=
p
2
m
e
E
e
.
We need to convert eV to joules(J).
E
e
(
J
) = 1000
eV
·
1
.
6
×
10

19
J
eV
= 1
.
6
×
10

16
J
p
e
=
p
2
*
9
.
0
×
10

31
kg
·
1
.
6
×
10

16
J
= 1
.
7
×
10

23
kg
m
s
.
Now we use the de Broglie relation to ﬁnd
λ
e
,
λ
e
=
h
p
e
=
6
.
6
×
10

34
J
·
s
1
.
7
×
10

23
kg
m
s
= 3
.
9
×
10

11
m
= 39
pm
= 0
.
039
nm.
3 Question 3
State the Pauli exclusion priciple.
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Unformatted text preview: The Pauli exclusion principle states that no two fermions can exist in the same quantum state . In the case of the atom, no two electrons can share the same 4 quantum numbers . Examples of fermions are, • electrons • protons • quarks All fermions share the same property of 1 2 interger spin(interges being 1 , 2 , 3 ... so on. Neutrons are not fermions since they do not have 1 2 interger spin. 1...
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This note was uploaded on 05/18/2008 for the course PHY 102 taught by Professor Simoncatterall during the Spring '08 term at Syracuse.
 Spring '08
 SimonCatterall
 Energy, Momentum, Photon

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