1.
CHAPTER 5:
Spectral Lines of
Hydrogen
(2 Hours)
DR.ATAR @ UiTM.NS
Dr Ahmad Taufek Abdul Rahman
PHY310 Spectral Lines of Hydrogen
School of Physics & Material Studies
Faculty of Applied Sciences
Universiti Teknologi MARA Malaysia
Campus of Negeri Sembilan
72000 Kuala Pilah, NS
1
2.
Learning Outcome:
5.1 Bohr‟s atomic model (1 hour)
At the end of this chapter, students should be able to:
Explain Bohr’s postulates of hydrogen atom.
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3.
5.1 Bohr‟s atomic model
5.1.1
Early models of atom
Thomson’s model of atom
In 1898, Joseph John Thomson suggested a model of an atom that consists
of homogenous positively charged spheres with tiny negatively charged
electrons embedded throughout the sphere as shown in Figure 1.
positively charged
sphere
electron
Figure 1
The electrons much likes currants in a plum pudding.
This model of the atom is called „plum pudding‟ model of the atom.
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PHY310 Spectral Lines of Hydrogen
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4.
Rutherford’s model of atom
In 1911, Ernest Rutherford performed a critical experiment that
showed the Thomson‟s model is not correct and proposed his new
atomic model known as Rutherford‟s planetary model of the atom as
shown in Figure 2.
nucleus
electron
Figure 2
According to Rutherford‟s model, the atom was pictured as
electrons orbiting around a central nucleus which concentrated of
positive charge.
The electrons are accelerating because their directions are
constantly changing as they circle the nucleus.
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PHY310 Spectral Lines of Hydrogen
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5.
Based on the wave theory, an accelerating charge emits energy.
Hence the electrons must emit the EM radiation as they revolve
around the nucleus.
As a result of the continuous loss of energy, the radii of the
electron orbits will be decreased steadily.
This would lead the electrons spiral and falls into the nucleus,
hence the atom would collapse as shown in Figure 3.
e
+Ze
„plop‟
energy loss
Figure 3
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6.
5.1.2 Bohr’s model of hydrogen atom
In 1913, Neils Bohr proposed a new atomic model based on
hydrogen atom.
According to Bohr‟s Model, he assumes that each electron
moves in a circular orbit which is centred on the nucleus,
the necessary centripetal force being provided by the
electrostatic force of attraction between the positively
charged nucleus and the negatively charged electron as
shown in Figure 4.
e
Fe
+e
v
Figure 4
r
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7.
On this basis he was able to show that the energy of an orbiting
electron depends on the radius of its orbit.
This model has several features which are described by the
postulates (assumptions) stated below :
1. The electrons move only in certain circular orbits, called
STATIONARY STATES or ENERGY LEVELS. When it is in
one of these orbits, it does not radiate energy.
2. The only permissible orbits are those in the discrete set for
which the angular momentum of the electron L equals an
integer times h/2π . Mathematically,
nh
L
2
nh
mvr
2
where
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and
L mvr
(1)
n : principalquantum number 1,2,3,...
r : radius of theorbit
m : mass of theelectron
PHY310 Spectral Lines of Hydrogen
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8.
3. Emission or absorption of radiation occurs only when
an electron makes a transition from one orbit to
another.
The frequency f of the emitted (absorbed) radiation is
given by
(2)
E hf Ef Ei
where
Note:
E : change of energy
h : Planck's constant
Ef : final energy state
Ei : initial energy state
If Ef > Ei
Absorption of EM radiation
If Ef < Ei
Emission of EM radiation
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9.
Learning Outcome:
5.2 Energy level of hydrogen atom (1 hour)
At the end of this chapter, students should be able to:
Derive Bohr’s radius and energy level in hydrogen atom.
Use
2
h
rn n a0 n
2
2
4 mk e
2
and
k e2 1
En
2
2 a0 n
Define ground state energy, excitation energy and ionisation
energy.
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PHY310 Spectral Lines of Hydrogen
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10.
5.2 Energy level of hydrogen atom
5.2.1 Bohr’s radius in hydrogen atom
Consider one electron of charge –e and mass m moves in a
circular orbit of radius r around a positively charged nucleus with
a velocity v as shown in Figure 11.3.
The electrostatic force between electron and nucleus
contributes the centripetal force as write in the relation below:
Fe Fc centripetal force
1 Q1Q2 mv 2
and Q1 Q2 e
2
40 r
r
e2
(3)
mv 2
40 r
electrostatic force
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PHY310 Spectral Lines of Hydrogen
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11.
From the Bohr‟s second postulate:
nh
mvr
2
By taking square of both side of the equation, we get
n2h2
m 2v 2 r 2
4 2
(4)
By dividing the eqs. (11.4) and (11.3), thus
n2h2
4 2
2 2 2
m v r
2
e
mv 2
4 r
0
n 2 h 2 0
r
me 2
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and
1
0
4k
PHY310 Spectral Lines of Hydrogen
electrostatic
constant
11
12.
n2h2
r
me 2
1
4k
2
h2
rn n
4 2 mk e2
; n 1,2,3...
(5)
which rn is radii of the permissible orbits for the Bohr‟s
atom.
Eq. (5) can also be written as
rn n a0
2
(6)
and
h2
a0
4 2 mk e2
where a0 is called the Bohr’s radius of hydrogen atom.
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13.
The Bohr‟s radius is defined as the radius of the most stable
(lowest) orbit or ground state (n=1) in the hydrogen atom
and its value is
a0
6.63 10
9.1110 9.00 10 1.60 10
34 2
4
2
31
a0 5.31 10
11
m
OR
9
19 2
0.531 Å (angstrom)
Unit conversion:
1 Å = 1.00 1010 m
The radii of the orbits associated with allowed orbits or states
n = 2,3,… are 4a0,9a0,…, thus the orbit’s radii are
quantized.
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14.
5.2.2 Energy level in hydrogen atom
is defined as a fixed energy corresponding to the orbits in
which its electrons move around the nucleus.
The energy levels of atoms are quantized.
The total energy level E of the hydrogen atom is given by
E U K
Potential energy of the electron
(7)
Kinetic energy of the electron
Potential energy U of the electron is given by
kQ1Q2
U
r
k e2
U 2
n a0
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where
Q1 e; Q2 e and r n 2 a0
nucleus electron
(8)
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15.
Kinetic energy K of the electron is given by
1 2
e2
K mv but mv 2
2
40 r
1 e2
where 1 k and r n 2 a0
K
2 40 r
40
1 ke2
(9)
K 2
n a
2
0
Therefore the eq. (11.7) can be written as
2
2
ke
1 ke
En 2 2
n a 0 2 n a0
ke2 1
En
2
2 a0 n
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PHY310 Spectral Lines of Hydrogen
(10)
15
16.
In general, the total energy level E for the atom is
ke2 Z 2
2
En
2 a0 n
where Z : atomic number
(11)
Using numerical value of k, e and a0, thus the eq. (11.10) can
2
be written as 9.00 10 9 1.60 10 19 1
En
Note:
11
2
n
2 5.31 10
2.17 10 18 1
eV 2
19
1.60 10
n
13.6
En 2 eV; n 1,2,3,...
(12)
n
th
where En : energy level of n state (orbit)
Eqs. (10) and (12) are valid for energy level of the hydrogen atom.
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17.
The negative sign in the eq. (11.12) indicates that work has to
be done to remove the electron from the bound of the atom
to infinity, where it is considered to have zero energy.
The energy levels of the hydrogen atom are when
n=1, the ground state (the state of the lowest energy level) ;
E1
13.6
1
2
eV 13.6 eV
n=2, the first excited state;
n=3, the second excited state;
n=4, the third excited state;
n=, the energy level is
E
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13.6
2
eV 0
E2
13.6
E3
13.6
2
2
E4
eV 3.40 eV
3
2
13.6
4
2
eV 1.51 eV
eV 0.85 eV
electron is completely
removed from the atom.
PHY310 Spectral Lines of Hydrogen
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18.
Figure 4 shows diagrammatically the various energy levels in the
hydrogen atom.
En (eV )
0.0 Free electron
5
4
3
th
0.54 4rd excited state
0.85 3 excited state
1.51 2nd excited state
n
Figure 4
Ionization energy
is defined as the
energy required by
an electron in the 2
ground state to
escape completely
from the attraction
of the nucleus.
An atom
becomes ion.
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1
3.40
1st
excited state
Excitation energy
is defined as the energy
required by an electron that
raises it to an excited state
from its ground state.
13.6 Ground state
PHY310 Spectral Lines of Hydrogen
excited state
is defined as the
energy levels
that higher
than the
ground state.
is defined as the
lowest stable
energy state of an
atom.
18
19.
Example 1 :
The electron in the hydrogen atom makes a transition from the
energy state of 0.54 eV to the energy state of 3.40 eV. Calculate
the wavelength of the emitted photon.
(Given the speed of light in the vacuum, c =3.00108 m s1 and
Planck‟s constant, h =6.631034 J s)
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PHY310 Spectral Lines of Hydrogen
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20.
Solution :
The change of the energy state in joule is given by
Ei 0.54 eV; Ef 3.40 eV
E Ef Ei
E 3.40 0.54
2.86 1.60 10 19
E 4.58 10 19 J
Therefore the wavelength of the emitted photon is
E
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hc
4.58 10 19
6.63 10 3.00 10
34
8
4.34 10 m
7
PHY310 Spectral Lines of Hydrogen
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21.
Example 2 :
The lowest energy state for hydrogen atom is 13.6 eV. Determine
the frequency of the photon required to ionize the atom.
(Given the speed of light in the vacuum, c =3.00108 m s1 and
Planck‟s constant, h =6.631034 J s)
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PHY310 Spectral Lines of Hydrogen
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22.
Solution :
The ionization energy in joule is given by
Ei 13.6 eV; Ef E 0
E Ef Ei
E 0 13.6
13.6 1.60 10 19
E 2.18 10 18 J
Therefore the frequency of the photon required to ionize the atom is
E hf
2.18 10 18 6.63 10 34 f
f 3.29 1015 Hz
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23.
Example 3 :
For an electron in a hydrogen atom characterized by the principal
quantum number n=2, calculate
a. the orbital radius,
b. the speed,
c. the kinetic energy.
(Given c =3.00108 m s1, h =6.631034 J s, me=9.111031 kg;
e=1.601019 C and k=9.00109 N m2 C2)
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PHY310 Spectral Lines of Hydrogen
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24.
Solution :
a. The orbital radius of the electron in the hydrogen atom for n=2
level is given by
n2
h2
2
rn n 2
4 mke2
34 2
6.63 10
2
r2 2
4 2 9.11 10 31 9.00 10 9 1.60 10 19
r2 2.12 10 10 m
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PHY310 Spectral Lines of Hydrogen
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24
25.
Solution : n 2
b. By applying the Bohr‟s 2nd postulate, thus
nh
mvrn
2
2h
mvr2
2
6.63 10 34
9.11 10 31 v 2.12 10 10
6
v 1.09 10 m s 1
c. The kinetic energy of the orbiting electron is given by
1 2
K mv
2
1
9.11 10 31 1.09 10 6
2
K 5.41 10 19 J
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PHY310 Spectral Lines of Hydrogen
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25
26.
Example 4 :
A hydrogen atom emits radiation of wavelengths 221.5 nm and 202.4
nm when the electrons make transitions from the 1st excited state
and 2nd excited state respectively to the ground state.
Calculate
a. the energy of a photon for each of the wavelengths above,
b. the wavelength emitted by the photon when the electron makes a
transition from the 2nd excited state to the 1st excited state.
(Given the speed of light in the vacuum, c =3.00108 m s1 and
Planck‟s constant, h =6.631034 J s)
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PHY310 Spectral Lines of Hydrogen
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27.
Solution :
a. The energy of the photon due to transition from 1st excited state
to the ground state is
1 221.5 10 9 m; 2 202.4 10 9 m
E1
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hc
1
E1
6.63 10 3.00 10
34
8
221 .5 10 9
E1 8.98 10 19 J
PHY310 Spectral Lines of Hydrogen
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28.
Solution :
1 221.5 10 9 m; 2 202.4 10 9 m
a. The energy of the photon due to transition from 2nd excited state
to the ground state is
E2
6.63 10 3.00 10
34
8
202.4 10 9
E2 9.83 10 19 J
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29.
1 221.5 10 9 m; 2 202.4 10 9 m
Solution :
b.
ΔE3
2nd excited state
1st excited state
ΔE1
ΔE2
E3 E2 E1
Ground state
E3 9.83 10 19 8.98 10 19
E3 8.50 10 20 J
Therefore the wavelength of the emitted photon due to the transition from 2nd
excited state to the 1st excited state is
E3
hc
3
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8.50 10 20
6.63 10 3.00 10
34
8
3 2.34 10 m
6 3
PHY310 Spectral Lines of Hydrogen
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30.
Learning Outcome:
5.3 Line spectrum (1 hour)
At the end of this chapter, students should be able to:
Explain the emission of line spectrum by using energy
level diagram.
State the line series of hydrogen spectrum.
Use formula,
E
hc
1
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31.
5.3 Line spectrum
The emission lines correspond to the photons of discrete
energies that are emitted when excited atomic states in the
gas make transitions back to lower energy levels.
Figure 11.5 shows line spectra produced by emission in the
visible range for hydrogen (H), mercury (Hg) and neon (Ne).
Figure 5
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32.
5.3.1 Hydrogen emission line spectrum
Emission processes in hydrogen give rise to series, which are
sequences of lines corresponding to atomic transitions.
The series in the hydrogen emission line spectrum are
Lyman series involves electron transitions that end at the
ground state of hydrogen atom. It is in the ultraviolet (UV)
range.
Balmer series involves electron transitions that end at the
1st excited state of hydrogen atom. It is in the visible light
range.
Paschen series involves electron transitions that end at
the 2nd excited state of hydrogen atom. It is in the infrared
(IR) range.
Brackett series involves electron transitions that end at the
3rd excited state of hydrogen atom. It is in the IR range.
Pfund series involves electron transitions that end at the
4th excited state of hydrogen atom. It is in the IR range.
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33.
Figure 6 shows diagrammatically the series of hydrogen
emission line spectrum.
En (eV )
0.0 Free electron
n
5
4
3
0.54 th
Pfund series0.85 4rd excited state
3 excited state
Brackett series
1.51 2nd excited state
Paschen series
2
3.39 1st excited state
Balmer series
Figure 6
Lyman series
1
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13.6 Ground state
PHY310 Spectral Lines of Hydrogen
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34.
Figure 7 shows “permitted” orbits of an electron in the Bohr
model of a hydrogen atom.
Figure 7: not to scale
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35.
5.3.2 Wavelength of hydrogen emission line spectrum
If an electron makes a transition from an outer orbit of level ni to
an inner orbit of level nf, thus the energy is radiated.
The energy radiated in form of EM radiation (photon) where
the wavelength is given by
E
E
hc
hc
1
(13)
From the Bohr‟s 3rd postulate, the eq. (11.13) can be written as
1
1
Enf Eni
hc
where
and
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ke2 1
Enf
2a0 nf 2
ke2 1
Eni
2
2a0 ni
PHY310 Spectral Lines of Hydrogen
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36.
1
ke2 1 ke2 1
2a0 nf 2 2a0 ni 2
1
ke2 1
1
2
hc
2a0 nf 2 ni
k e2 1
1
ke2
2 and
RH
n 2 n
2hca0 f
2hca0
i
1
hc
1
1
RH 2 2
n
ni
f
1
where
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(14)
RH : Rydberd' s constant 1.097107 m1
nf : final value of n
ni : initial value of n
PHY310 Spectral Lines of Hydrogen
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37.
Note:
For the hydrogen line spectrum,
Lyman series( nf=1 )
Balmer series( nf=2 )
1
1
RH 2 2
1 n
i
1
1
1
RH 2 2
2
ni
1
1
1
RH 2 2
3
ni
1
1
1
RH 2 2
4
ni
1
1
1
RH 2 2
5
ni
1
Brackett series( nf=4 )
Paschen series( nf=3 )
Pfund series( nf=5 )
To calculate the shortest wavelength in any series, take ni= .
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38.
5.3.3 Limitation of Bohr’s model of hydrogen atom
The Bohr‟s model of hydrogen atom
predicts successfully the energy levels of the hydrogen atom but
fails to explain the energy levels of more complex atoms.
can explain the spectrum for hydrogen atom but some details of
the spectrum cannot be explained especially when the atom
is placed in a magnetic field.
cannot explain the Zeeman effect (Figure 11.7).
Zeeman effect is defined as the splitting of spectral lines
when the radiating atoms are placed in a magnetic field.
2
1
Transitio
ns
No magnetic
field
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Energy
Levels
Magnetic
field
Figure 7
Spectra
PHY310 Spectral Lines of Hydrogen
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39.
Example 5 :
The Balmer series for the hydrogen atom corresponds to electronic transitions
that terminate at energy level n=2 as shown in the Figure 8.
6
5
4
3
En (eV )
0 .0
0.38
0.54
0.85
1.51
2
3.40
n
Figure 8
Calculate
a. the longest wavelength, and
b. the shortest wavelength of the photon emitted in this series.
(Given the speed of light in the vacuum c =3.00108 m s1 ,Planck‟s constant h
=6.631034 J s and Rydberg‟s constant RH = 1.097 107 m1)
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PHY310 Spectral Lines of Hydrogen
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40.
Solution : nf 2
a. The longest wavelength of the photon results due to the
electron transition from n = 3 to n = 2 (Balmer series). Thus
1
1
1
E
RH 2 2
OR
n
ni
hc
f
1 Ef Ei
1
1
7 1
1.097 10 2 2
hc
max
3
2
E 2 E3
1
max 6.56 10 7 m
max
hc
3.40 1.51 1.60 10 19
6.63 10 34 3.00 108
1
max 6.58 10 7 m
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PHY310 Spectral Lines of Hydrogen
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41.
Solution : n 2
f
b. The shortest wavelength of the photon results due to the electron
transition from n = to n = 2 (Balmer series). Thus
1
Ef Ei
hc
1
min
E 2 E
hc
3.40 0 1.60 10 19
6.63 10
34
min 3.66 10 7 m
3.00 10
8
1
1
OR
RH 2 2
n
ni
f
1
1
7 1
1.097 10 2 2
min
2
1
min 3.65 10 7 m
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PHY310 Spectral Lines of Hydrogen
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42.
Example 6 :
Determine the wavelength for a line spectrum in Lyman series
when the electron makes a transition from n=3 level.
(Given Rydberg‟s constant ,RH = 1.097 107 m1)
Solution :
ni 3 ; nf 1
By applying the equation of wavelength for Lyman series, thus
1
1
RH 2 2
1 n
i
1
7 1
1.097 10 2 2
1 3
1
1.03 10 7 m
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43.
Exercise 5.1 :
Given c =3.00108 m s1, h =6.631034 J s, me=9.111031 kg,
e=1.601019 C and RH =1.097107 m1
1.
A hydrogen atom in its ground state is excited to the n =5
level. It then makes a transition directly to the n =2 level
before returning to the ground state. What are the
wavelengths of the emitted photons?
ANS. :
4.34107 m; 1.22107 m
2.
Show that the speeds of an electron in the Bohr orbits are
given ( to two significant figures) by
vn
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2.2 10
6
m s 1
n
PHY310 Spectral Lines of Hydrogen
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