Test # 3 Solutions
1(a) The ground state of the Hydrogen atom (
n
= 1) is
E
1
=

13
.
6eV. Thus,
we require exactly 13
.
6eV to ionize the Hydrogen atom in its ground state.
1(b) With
E
initial
= 0 eV and
E
final
=

3
.
4 eV, we have
E
=
hf
=
E
initial

E
final
= 0

(

3
.
4) = 3
.
4 eV. Then, using the fact that
h
= 4
.
14
×
10

15
eV s,
and that
f
=
c/λ
where
c
= 3
.
0
×
10
8
m s

1
is the speed of light, we get
c
λ
=
E
initial

E
final
= 3
.
4 eV
⇒
λ
=
hc
E
=
4
.
14
×
10

15
eV s
×
3
.
0
×
10
8
m s

1
3
.
4eV
∼
=
3
.
65
×
10

7
m
Thus, the emitted photon has a wavelength of nearly 3
.
65
×
10

7
m.
We may also use the alternative approach, where we use the formula
1
λ
=
R
(
1
n
2

1
m
2
)
In the above expression,
R
= 1
.
097
×
10
7
m

1
.
However, we now need to be
careful, since we are not starting off with the Hydrogen atom. Instead, we are
starting out with a single proton, to which an electron is brought so that they
together form a Hydrogen atom in the first excited state (
n
= 2). But initially,
we were not in the Hydrogen atom at all, so the proper value for
m
in this setup
would be
m
=
∞
for which (1
/m
) = (1
/m
2
) = 0. You can check for yourself
that if we plug these values above, we get the same answer,
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 Spring '08
 SAI
 Physics, Atom, Light, #, Fundamental physics concepts, 1.3m

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