Question

What is the ionization energy of a hydrogen atom that is in the n = 6 excited state?

Answer 1

Answer:
36.46

kJ/mol

Explanation:
The Rydberg Expression is given by:
1
λ
=
R
[
1
n
2
1
−
1
n
2
2
]

λ
is the wavelength of the emission line
n
1
is the principle quantum number of the lower energy level
n
2
is the principle quantum number of the higher energy level
R
is the Rydberg Constant
1.097
×
10
7

m
−
1




At higher and higher values of
n
2
the term
1
n
2
2
tends to zero. Effectively
n
2
=
∞
and the electron has left the atom, forming a hydrogen ion.
The Rydberg Expression refers to emission where an electron falls from a higher energy level to a lower one, emitting a photon.
In this case we can use it to find the energy required to move an electron from
n
=
6
to
n
=
∞
.
The expression now becomes:
1
λ
=
R
[
1
n
2
1
−
0
]

∴
1
λ
=
R
n
2
1

Since
n
1
=
6
this becomes:
1
λ
=
R
36

∴
λ
=
36
R
=
36
1.097
×
10
7
=
32.816
×
10
−
7

m
To convert this into energy we use the Planck expression:
E
=
h
f
=
h
c
λ

∴
E
=
6.626
×
10
−
34
×
3
×
10
8
32.816
×
10
−
7
=
6.0575
×
10
−
20
J
You'll notice from the graphic that the energy of the
n
=
6
electron is
−
0.38
eV
.
This means the energy to remove it will be
+
0.38
eV
.
To convert this to Joules you multiply by the electronic charge
=
0.38
×
1.6
×
10
−
19
=
6.08
×
10
−
20

J

As you can see my calculated value is very close to this.
This is the energy required to ionise a single atom. To get the energy required to ionise a mole of atoms we multiply by the Avogadro Constant:
E
=
6.0575
×
10
−
20
×
6.02
×
10
23

J/mol

E
=
36.461
×
10
3

J/mol

E
=
36.46

kJ/mol