Chapter 37
Atoms
λ
= (1240/
∆
E
) nm, where
∆
E
=
E
i
–
E
f
and the energies are in eV (see Equs. 17-5 and 17-21); here
E
f
= -13.6 eV.
(
a
)
E
3
= –13.6/9 eV = –1.51 eV
(
b
)
E
4
= –13.6/16 eV = –0.85 eV
∆
E
= 12.09 eV;
= 102.6 nm
∆
E
= 12.75 eV;
= 97.25 nm
9*
∙
Find the photon energy for the three longest wavelengths in the Balmer series and calculate the
wavelengths.
For the Balmer series,
E
f
=
E
(
n
= 2) = –3.40 eV. Use Equs. 17-5 and 17-21, i.e.,
= (1240 eV
.
nm)/(
∆
E
eV).
1.
∆
E
=
E
3
–
E
2
;
E
3
= –13.6/9 eV = –1.51 eV
2.
∆
E
=
E
4
–
E
2
;
E
4
= –13.6/16 eV = –0.85 eV
3.
∆
E
=
E
5
–
E
2
;
E
5
= –13.6/25 eV = –0.544 eV
∆
E
= 1.89 eV;
3,2
= 656.1 nm
∆
E
= 2.55 eV;
4,2
= 486.3 nm
∆
E
= 2.856 eV;
5,2
= 434.2 nm
10
∙
(
a
) Find the photon energy and wavelength for the series limit (shortest wavelength) in the Paschen series
(
n
2
= 3). (
b
) Calculate the wavelengths for the three longest wavelengths in this series and indicate their
positions on a horizontal linear scale.
(
a
) For
min
,
n
=
∞
and
E
i
= 0
h
_
f
= (13.6/9) eV = 1.51 eV;
min
= 1240/1.51 nm = 820.6 nm
(
b
) For the three longest
,
n
i
= 4, 5, and 6
∆
E
4,3
= 0.66 eV;
4,3
= 1876 nm
∆
E
5,3
= 0.967 eV;
5,3
= 1282 nm
∆
E
6,3
= 1.133 eV;
6,3
= 1094 nm
The positions of these lines on a horizontal linear scale are shown below with the wavelengths indicated.
11
∙
Repeat Problem 10 for the Brackett series (
n
2
= 4).
(
a
) For
min
,
n
=
∞
and
E
i
= 0
(
b
) For the three longest
,
n
i
= 5,
6, and 7
h
_
f
= (13.6/16) eV = 0.85 eV;
λ
min
= (1240/0.85) nm = 1459 nm
∆
E
5,4
= 0.306 eV;
4,3
= 4052 nm
∆
E
6,4
= 0.472 eV;
5,3
= 2627 nm
∆
E
7,4
= 0.572 eV;
6,3
= 2166 nm
The positions of these lines on a horizontal linear scale are shown below with the wavelengths indicated.