Embed Size (px)
Citation preview
1
PHYSICS CHAPTER 11
CHAPTER 11: CHAPTER 11: Bohr’s model of hydrogen atomBohr’s model of hydrogen atom
(3 Hours)(3 Hours)
PHYSICS CHAPTER 11
2
At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: ExplainExplain Bohr’s postulates of hydrogen atom. Bohr’s postulates of hydrogen atom.
Learning Outcome:
ww
w.k
mp
h.m
atri
k.ed
u.m
y/p
hys
ics
ww
w.k
mp
h.m
atri
k.ed
u.m
y/p
hys
ics
11.1 Bohr’s atomic model (1 hour)
PHYSICS CHAPTER 11
3
11.1.1 Early models of atomThomson’s model of atomThomson’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 11.1.
The electrons much likes currants in a plum pudding. This model of the atom is called ‘plum pudding’ model of the
atom.
11.1 Bohr’s atomic model
positively charged sphere
electron
Figure 11.1Figure 11.1
PHYSICS CHAPTER 11
4
Rutherford’s model of atomRutherford’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 11.2a.
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.
nucleus electron
Figure 11.2aFigure 11.2a
PHYSICS CHAPTER 11
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 11.2b.
Figure 11.2bFigure 11.2b
++Ze ee
‘plop’
energy loss
PHYSICS CHAPTER 11
6
+e
e
v
r
eF
In 1913, Neils Bohr proposed a new atomic model based on hydrogen atom.
According to Bohr’s Model, he assumes that each electron each electron moves in a circular orbit which is centred on the nucleusmoves in a circular orbit which is centred on the nucleus, the necessary centripetal force being provided by the centripetal force being provided by the electrostatic force of attraction between the positively electrostatic force of attraction between the positively charged nucleus and the negatively charged electroncharged nucleus and the negatively charged electron as shown in Figure 11.3.
11.1.2 Bohr’s model of hydrogen atom
Figure 11.3Figure 11.3
PHYSICS CHAPTER 11
7
On this basis he was able to show that the energy of an energy of an orbiting electron depends on the radius of its orbitorbiting electron depends on the radius of its orbit.
This model has several features which are described by the postulates (assumptions) stated below :
1. The electronselectrons move only in certain circular orbits, called STATIONARY STATESSTATIONARY STATES or ENERGY LEVELSENERGY LEVELS. When it is in one of these orbits, it does not radiate energydoes not radiate energy.
2. The only permissible orbits arepermissible orbits are those in the discrete set for which the angular momentum of the electron angular momentum of the electron LL
equals an integer times equals an integer times h/h/22ππ . Mathematically,
2
nhL
2
nhmvr (11.1)(11.1)
and mvrL
whereorbit theof radius: r
electron theof mass:m
,...,,n 321number quantum principal:
PHYSICS CHAPTER 11
8
3. Emission or absorptionEmission or absorption of radiation occurs only when an electron makes a transition from one orbit to anotherelectron makes a transition from one orbit to another.
The frequency f of the emitted (absorbed) radiation is given by
if EEhfE (11.2)(11.2)
whereconstant sPlanck': h
stateenergy final:fE
energy of change: E
stateenergy initial:iENote:Note:
If Ef > Ei
If Ef < EiEmissionEmission of EM radiation
AbsorptionAbsorption of EM radiation
PHYSICS CHAPTER 11
9
At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: DeriveDerive Bohr’s radius and energy level in hydrogen atom. Bohr’s radius and energy level in hydrogen atom. UseUse
DefineDefine ground state energy, excitation energy and ground state energy, excitation energy and ionisation energy.ionisation energy.
Learning Outcome:
ww
w.k
mp
h.m
atri
k.ed
u.m
y/p
hys
ics
ww
w.k
mp
h.m
atri
k.ed
u.m
y/p
hys
ics
11.2 Energy level of hydrogen atom (1 hour)
222
02
4 mke
hnanrn
andand
20
2 1
2 na
keEn
PHYSICS CHAPTER 11
10
11.2.1 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 electrostatic force between electron and nucleus
contributes the centripetal forcecontributes the centripetal force as write in the relation below:
11.2 Energy level of hydrogen atom
ce FF centripetal forceelectrostatic force
r
mv
r
QQ 2
221
04
1
and eQQ 21
r
emv
0
22
4 (11.3)(11.3)
PHYSICS CHAPTER 11
11
From the Bohr’s second postulate:
By taking square of both side of the equation, we get
By dividing the eqs. (11.4) and (11.3), thus
2
nhmvr
(11.4)(11.4)2
22222
4hn
rvm
r
e
hn
mv
rvm
0
2
2
22
2
222
4
4
20
22
me
hnr and
k
4
10
electrostatic electrostatic constantconstant
PHYSICS CHAPTER 11
12
which rn is radii of the permissible orbitsradii of the permissible orbits for the Bohr’s atom. Eq. (11.5) can also be written as
where a0 is called the Bohr’s radiusBohr’s radius of hydrogen atom.
kme
hnr
4
12
22
(11.5)(11.5)...3,2,1;4 22
22
n
mke
hnrn
02anrn
22
2
04 mke
ha
(11.6)(11.6)
and
PHYSICS CHAPTER 11
13
The Bohr’s radius is defined as the radius of the most stable the radius of the most stable (lowest) orbit or ground state ((lowest) orbit or ground state (nn=1=1)) in the hydrogen atomin the hydrogen atom and its value is
Unit conversion:
The radii of the orbits associated with allowed orbits or states
n = 2,3,… are 4a0,9a0,…, thus the orbit’s radii are orbit’s radii are quantizedquantized.
2199312
234
01060.11000.91011.94
1063.6
a
m 1031.5 110
a OR 0.531 Å (angstrom)
1 Å = 1.00 1010 m
PHYSICS CHAPTER 11
14
is defined as a fixed energy corresponding to the orbits in a fixed energy corresponding to the orbits in which its electrons move around the nucleuswhich its electrons move around the nucleus.
The energy levels of atoms are quantizedquantized. The total energy level total energy level EE of the hydrogen atom of the hydrogen atom is given by
Potential energy Potential energy UU of the electron of the electron is given by
11.2.2 Energy level in hydrogen atom
KUE (11.7)(11.7)
Kinetic energy of the electronKinetic energy of the electronPotential energy of the electronPotential energy of the electron
r
QkQU 21 eQeQ 21 ;where 0
2anr and
02
2
an
keU (11.8)(11.8)
nucleusnucleus electronelectron
PHYSICS CHAPTER 11
15
Kinetic energy Kinetic energy KK of the electron of the electron is given by
Therefore the eq. (11.7) can be written as
2
2
1mvK
(11.9)(11.9)
butr
emv
0
22
4
r
eK
0
2
42
1
where k
04
1
02
2
2
1
an
keK
02
2
02
2
2
1
an
ke
an
keEn
and 02anr
20
2 1
2 na
keEn (11.10)(11.10)
PHYSICS CHAPTER 11
16
In general, the total energy level E for the atom is
Using numerical value of k, e and a0, thus the eq. (11.10) can
be written as
2
2
0
2
2 n
Z
a
keEn (11.11)(11.11)
211
2199 1
1031.52
1060.11000.9
nEn
219
18 1eV
1060.1
1017.2
n
1,2,3,... eV; 6.13
2 n
nEn (11.12)(11.12)
Note:Note:
Eqs. (11.10) and (11.12) are valid for energy level of the hydrogen atomEqs. (11.10) and (11.12) are valid for energy level of the hydrogen atom.
where number atomic :Z
where (orbit) state of levelenergy : thnEn
PHYSICS CHAPTER 11
17
The negative signnegative sign in the eq. (11.12) indicates that work has to work has to be done to remove the electron from the bound of the atom be done to remove the electron from the bound of the atom to infinityto infinity, where it is considered to have zero energyzero energy.
The energy levels of the hydrogen atom are when
n=1, the ground stateground state (the state of the lowest energy levellowest energy level) ;
n=2, the first excited statefirst excited state;
n=3, the second excited statesecond excited state;
n=4, the third excited statethird excited state;
n=, the energy level is
eV 613eV
1
6.1321 .E
eV 403eV
2
6.1322 .E
0eV
6.132
E
eV 511eV
3
6.1323 .E
eV 850eV
4
6.1324 .E
electron is completely electron is completely removed from the atomremoved from the atom.
PHYSICS CHAPTER 11
18
Figure 11.4 shows diagrammatically the various energy levels in the hydrogen atom.
excited stateexcited state
is defined as the the lowest stable lowest stable energy state of energy state of an atom.an atom.
is defined as the energy the energy levels that levels that higher than higher than the ground the ground state.state.
)(eVEnn 0.0
5 54.04 85.03 51.1
2 40.3
1 6.13
Excitation energyExcitation energyis defined as the energy the energy required by an electron that required by an electron that raises it to an excited state raises it to an excited state from its ground state.from its ground state.
Ionization energyIonization energyis defined as the the energy required energy required by an electron in by an electron in the ground state the ground state to escape to escape completely from completely from the attraction of the attraction of the nucleus.the nucleus.
An atom becomes ion. Ground state
1st excited state
2nd excited state3rd excited state4th excited state
Free electronFigure 11.4Figure 11.4
PHYSICS CHAPTER 11
19
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)
Solution :Solution :
The change of the energy state in joule is given by
Therefore the wavelength of the emitted photon is
Example 1 :
eV 40.3eV; 54.0 fi EE
if EEE 54.040.3 E
191060.186.2 J 1058.4 19E
hc
E
83419 1000.31063.6
1058.4
m 1034.4 7
PHYSICS CHAPTER 11
20
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)
Solution :Solution :
The ionization energy in joule is given by
Therefore the frequency of the photon required to ionize the atom is
Example 2 :
0eV; 6.13 fi EEE
if EEE 6.130 E
191060.16.13 J 1018.2 18E
hfE
f3418 1063.61018.2
Hz 1029.3 15f
PHYSICS CHAPTER 11
21
For an electron in a hydrogen atom characterized by the principal quantum number n=2, calculatea. 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)Solution :Solution :
a. The orbital radius of the electron in the hydrogen atom for n=2 level is given by
Example 3 :
2n
22
22
4 mke
hnrn
2199312
2342
21060.11000.91011.94
1063.62
r
m 1012.2 102
r
PHYSICS CHAPTER 11
22
Solution :Solution :
b. By applying the Bohr’s 2nd postulate, thus
c. The kinetic energy of the orbiting electron is given by
341031 1063.6
1012.21011.9
v
16 s m 1009.1 v
2
nhmvrn
2n
2
22
hmvr
2
2
1mvK
2631 1009.11011.92
1
J 1041.5 19K
PHYSICS CHAPTER 11
23
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.Calculatea. 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)
Solution :Solution :a. The energy of the photon due to transition from 1st excited state to the ground state is
Example 4 :
m 104.202m; 105.221 92
91
11
hcE
9
834
1105.221
1000.31063.6
E
J 1098.8 191
E
PHYSICS CHAPTER 11
24
Solution :Solution :a. The energy of the photon due to transition from 2nd excited state to the ground state is
b.
Therefore the wavelength of the emitted photon due to the transition from 2nd excited state to the 1st excited state is
m 104.202m; 105.221 92
91
9
834
2104.202
1000.31063.6
E
J 1083.9 192
E
ΔE1 ΔE2
ΔE3
Ground state
1st excited state
2nd excited state
123 EEE 19193 1098.81083.9 E
J 1050.8 203
E
33
hcE
3
83420 1000.31063.6
1050.8
m 1034.2 63
PHYSICS CHAPTER 11
25
At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: ExplainExplain the emission of line spectrum by using energy the emission of line spectrum by using energy
level diagram.level diagram. StateState the line series of hydrogen spectrum. the line series of hydrogen spectrum. UseUse formula, formula,
Learning Outcome:
ww
w.k
mp
h.m
atri
k.ed
u.m
y/p
hys
ics
ww
w.k
mp
h.m
atri
k.ed
u.m
y/p
hys
ics
11.3 Line spectrum (1 hour)
hc
E
1
PHYSICS CHAPTER 11
26
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).
11.3 Line spectrum
Figure 11.5Figure 11.5
PHYSICS CHAPTER 11
27
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 seriesLyman series involves electron transitions electron transitions that end at the end at the
ground state of hydrogen atomground state of hydrogen atom. It is in the ultraviolet ltraviolet (UV) range(UV) range.
Balmer seriesBalmer series involves electron transitions electron transitions that end at end at the 1the 1stst excited state of hydrogen atom excited state of hydrogen atom. It is in the visible visible light rangelight range.
Paschen seriesPaschen series involves electron transitions electron transitions that end at end at the 2the 2ndnd excited state of hydrogen atom excited state of hydrogen atom. It is in the infrared (IR) rangeinfrared (IR) range.
Brackett seriesBrackett series involves electron transitions electron transitions that end at end at the 3the 3rdrd excited state of hydrogen atom excited state of hydrogen atom. It is in the IR IR rangerange.
Pfund series Pfund series involves electron transitions electron transitions that end at the end at the 44thth excited state of hydrogen atom excited state of hydrogen atom. It is in the IR rangeIR range.
11.3.1 Hydrogen emission line spectrum
PHYSICS CHAPTER 11
28
Figure 11.6 shows diagrammatically the series of hydrogen emission line spectrum.
Figure 11.6Figure 11.6)eV(nE
0.0
54.085.051.1
39.3
6.13
n
43
2
1
5
Ground state
1st excited state
2nd excited state3rd excited state4th excited state
Free electron
Lyman seriesLyman series
Balmer seriesBalmer series
Paschen seriesPaschen seriesBrackett seriesBrackett series
Pfund seriesPfund series
Stimulation 11.1
PHYSICS CHAPTER 11
29
Figure 11.7 shows “permitted” orbits of an electron in the Bohr model of a hydrogen atom.
Figure 11.6: not to scaleFigure 11.6: not to scale
Picture 11.1
PHYSICS CHAPTER 11
30
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 radiatedenergy radiated in form of EM radiation (photon)form of EM radiation (photon) where the wavelength is given by
From the Bohr’s 3rd postulate, the eq. (11.13) can be written as
11.3.2 Wavelength of hydrogen emission line spectrum
hc
E hc
E
1
(11.13)(11.13)
if
11nn EE
hc
where
2f0
2 1
2f na
keEn
and
2i0
2 1
2i na
keEn
PHYSICS CHAPTER 11
31
2i0
2
2f0
2 1
2
1
2
11
na
ke
na
ke
hc
2i
2f0
2 11
2
1
nna
ke
hc
2i
2f0
2 11
2 nnhca
keand HR
hca
ke
0
2
2
2i
2f
111
nnRH
(11.14)(11.14)
where17 m 10097.1constant sRydberd': HR
nn of valuefinal: f
nn of valueinitial: i
PHYSICS CHAPTER 11
32
Note:Note:
For the hydrogen line spectrum,Lyman Lyman series( nnff=1=1 )
Balmer Balmer series( nnff=2=2 )
Paschen Paschen series( nnff=3=3 )
Brackett Brackett series( nnff=4=4 )
Pfund Pfund series( nnff=5=5 )
To calculate the shortest wavelength in any seriesshortest wavelength in any series, take nnii== .
2i
2
1
1
11
nRH
2i
2
1
2
11
nRH
2i
2
1
3
11
nRH
2i
2
1
4
11
nRH
2i
2
1
5
11
nRH
PHYSICS CHAPTER 11
33
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 fails to explain the energy levels of more complex atomsatoms.
can explain the spectrum for hydrogen atom but some details of the spectrum cannot be explained especially cannot be explained especially when the atom is placed in a magnetic fieldwhen the atom is placed in a magnetic field.
cannot explain the Zeeman effect (Figure 11.7). Zeeman effectZeeman effect is defined as the splitting of spectral the splitting of spectral
lines when the radiating atoms are placed in a lines when the radiating atoms are placed in a magnetic field.magnetic field.
11.3.3 Limitation of Bohr’s model of hydrogen atom
Magnetic field
Transitions
No magnetic field
11
22
Energy Levels
SpectraFigure 11.7Figure 11.7
PHYSICS CHAPTER 11
34
The Balmer series for the hydrogen atom corresponds to electronic transitions that terminate at energy level n=2 as shown in the Figure 11.8.
Calculatea. the longest wavelength, andb. 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)
Example 5 :
)eV(nE0.0
38.0
85.0
51.1
40.3
6
n
54
3
2
54.0
Figure 11.8Figure 11.8
PHYSICS CHAPTER 11
35
Solution :Solution :
a. The longest wavelength of the photon results due to the electron
transition from n = 3 to n = 2 (Balmer series). Thus
hc
EE if1
hc
E
1
hc
EE 32
max
1
834
19
1000.31063.6
1060.151.140.3
m 1058.6 7max
OR
2i
2f
111
nnRH
227
max 3
1
2
110097.1
1
m 1056.6 7max
2f n
PHYSICS CHAPTER 11
36
Solution :Solution :
b. The shortest wavelength of the photon results due to the electron
transition from n = to n = 2 (Balmer series). Thus
hc
EE 2
min
1
834
19
1000.31063.6
1060.1040.3
m 1066.3 7min
OR
2i
2f
111
nnRH
227
min
1
2
110097.1
1
m 1065.3 7
min
2f n
hc
EE if1
PHYSICS CHAPTER 11
37
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 :Solution :
By applying the equation of wavelength for Lyman series, thus
Example 6 :
1; 3 fi nn
2i
2
1
1
11
nRH
227
3
1
1
110097.1
m 1003.1 7
PHYSICS CHAPTER 11
38
Exercise 11.1 :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?
(College Physics, 6(College Physics, 6thth edition, Wilson, Buffa & Lou, Q66, p.875) edition, Wilson, Buffa & Lou, Q66, p.875)
ANS. :ANS. : 4.344.34101077 m; 1.22 m; 1.22101077 m m
2. Show that the speeds of an electron in the Bohr orbits are given ( to two significant figures) by
(College Physics, 6(College Physics, 6thth edition, Wilson, Buffa & Lou, Q66, p.875) edition, Wilson, Buffa & Lou, Q66, p.875)
n
vn
16 s m 102.2
39
PHYSICS CHAPTER 11
Next Chapter…CHAPTER 12 :
X-rays
