Phy 310 chapter 5

DR.ATAR @ UiTM.NS PHY310 Spectral Lines of Hydrogen 1

CHAPTER 5:

Spectral Lines of

Hydrogen

(2 Hours)

Dr Ahmad Taufek Abdul RahmanSchool of Physics & Material Studies

Faculty of Applied Sciences

Universiti Teknologi MARA Malaysia

Campus of Negeri Sembilan

72000 Kuala Pilah, NS


At the end of this chapter, students should be able to:

Explain Bohr’s postulates of hydrogen atom.

Learning Outcome:

5.1 Bohr‟s atomic model (1 hour)


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.

The electrons much likes currants in a plum pudding.

This model of the atom is called „plum pudding‟ model of the atom.

5.1 Bohr‟s atomic model

positively charged

sphere

electron

Figure 1


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.

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.

nucleuselectron

Figure 2


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.

Figure 3

+Ze e

„plop‟

energy loss


+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

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.

5.1.2 Bohr’s model of hydrogen atom

Figure 4


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 forwhich the angular momentum of the electron L equals aninteger times h/2π . Mathematically,

2

nhL

2

nhmvr (1)

and mvrL

where

orbit theof radius: relectron theof mass:m

,...,,n 321number quantum principal:


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

if EEhfE (2)

where

constant sPlanck': hstateenergy final:fE

energy of change: E

stateenergy initial:iENote:

If Ef > Ei

If Ef < Ei Emission of EM radiation

Absorption of EM radiation



Derive Bohr’s radius and energy level in hydrogen atom.

Use

Define ground state energy, excitation energy and ionisationenergy.

Learning Outcome:5.2 Energy level of hydrogen atom (1 hour)

22

20

2

4 mke

hnanrn

and

20

2 1

2 na

keEn


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:

5.2 Energy level of hydrogen atom

ce FF centripetal forceelectrostatic force

r

mv

r

QQ 2

2

21

04

1

and eQQ 21

r

emv

0

22

4 (3)


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

(4)2

22222

4

hnrvm

r

e

hn

mv

rvm

0

2

2

22

2

222

4

4

2

022

me

hnr and

k

4

10

electrostatic

constant


which rn is radii of the permissible orbits for the Bohr‟s

atom.

Eq. (5) can also be written as

where a0 is called the Bohr’s radius of hydrogen atom.

kme

hnr

4

12

22

(5)...3,2,1;4 22

22

n

mke

hnrn

02anrn

22

2

04 mke

ha

(6)

and


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

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

quantized.

2199312

234

0

1060.11000.91011.94

1063.6

a

m 1031.5 110

a OR 0.531 Å (angstrom)

1 Å = 1.00 1010 m


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

Potential energy U of the electron is given by

5.2.2 Energy level in hydrogen atom

KUE (7)

Kinetic energy of the electronPotential energy of the electron

r

QkQU 21 eQeQ 21 ;where

02anr and

02

2

an

keU (8)

nucleus electron


Kinetic energy K of the electron is given by

Therefore the eq. (11.7) can be written as

2

2

1mvK

(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 (10)


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)

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 (12)

Note:

Eqs. (10) and (12) are valid for energy level of the hydrogen atom.

where number atomic :Z

where (orbit) state of levelenergy : thnEn


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) ;

n=2, the first excited state;

n=3, the second excited state;

n=4, the third 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

removed from the atom.


Figure 4 shows diagrammatically the various energy levels in the hydrogen atom.

excited state

is defined as the

lowest stable

energy state of an

atom.

is defined as the

energy levels

that higher

than the

ground state.

)(eVEnn

0.0

5 54.04 85.0

3 51.1

2 40.3

1 6.13

Excitation energy

is defined as the energy

required by an electron that

raises it to an excited state

from its ground state.

Ionization energy

is defined as the

energy required by

an electron in the

ground state to

escape completely

from the attraction

of the nucleus.

An atom

becomes ion. Ground state

1st excited state

2nd excited state

3rd excited state4th excited state

Free electronFigure 4


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)

Example 1 :


Solution :

The change of the energy state in joule is given by

Therefore the wavelength of the emitted photon is

eV 40.3eV; 54.0 fi EE

if EEE 54.040.3 E

191060.186.2

J 1058.4 19E

hcE

83419 1000.31063.6

1058.4

m 1034.4 7


The lowest energy state for hydrogen atom is 13.6 eV. Determinethe frequency of the photon required to ionize the atom.



Example 2 :


Solution :

The ionization energy in joule is given by

Therefore the frequency of the photon required to ionize the atom is

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


For an electron in a hydrogen atom characterized by the principalquantum 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)

Example 3 :


Solution :

a. The orbital radius of the electron in the hydrogen atom for n=2

level is given by

2n

22

22

4 mke

hnrn

2199312

2342

2

1060.11000.91011.94

1063.62

r

m 1012.2 102

r


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


A hydrogen atom emits radiation of wavelengths 221.5 nm and 202.4nm when the electrons make transitions from the 1st excited stateand 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.



Example 4 :


Solution :

a. The energy of the photon due to transition from 1st excited state

to the ground state is

m 104.202m; 105.221 92

91

1

1

hcE

9

834

1105.221

1000.31063.6

E

J 1098.8 191

E


Solution :

a. The energy of the photon due to transition from 2nd excited state to the ground state is

m 104.202m; 105.221 92

91

9

834

2104.202

1000.31063.6

E

J 1083.9 192

E


Solution :

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

ΔE1 ΔE2

ΔE3

Ground state

1st excited state

2nd excited state

123 EEE 1919

3 1098.81083.9 EJ 1050.8 20

3E

3

3

hcE

3

83420 1000.31063.6

1050.8

m 1034.2 63



Explain the emission of line spectrum by using energy

level diagram.

State the line series of hydrogen spectrum.

Use formula,

Learning Outcome:

5.3 Line spectrum (1 hour)

hc

E

1


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).

5.3 Line spectrum

Figure 5


Emission processes in hydrogen give rise to series, which aresequences of lines corresponding to atomic transitions.

The series in the hydrogen emission line spectrum are

Lyman series involves electron transitions that end at theground state of hydrogen atom. It is in the ultraviolet (UV)range.

Balmer series involves electron transitions that end at the1st excited state of hydrogen atom. It is in the visible lightrange.

Paschen series involves electron transitions that end atthe 2nd excited state of hydrogen atom. It is in the infrared(IR) range.

Brackett series involves electron transitions that end at the3rd excited state of hydrogen atom. It is in the IR range.

Pfund series involves electron transitions that end at the4th excited state of hydrogen atom. It is in the IR range.

5.3.1 Hydrogen emission line spectrum


Figure 6 shows diagrammatically the series of hydrogen emission line spectrum.

Figure 6

)eV(nE0.0

54.085.051.1

39.3

6.13

n

43

2

1

5

Ground state

1st excited state

2nd excited state

3rd excited state4th excited state

Free electron

Lyman series

Balmer series

Paschen series

Brackett series

Pfund series


Figure 7 shows “permitted” orbits of an electron in the Bohrmodel of a hydrogen atom.

Figure 7: not to scale


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

From the Bohr‟s 3rd postulate, the eq. (11.13) can be written as

5.3.2 Wavelength of hydrogen emission line spectrum

hcE

hc

E

1(13)

if

11nn EE

hc

where

2

f0

2 1

2fna

keEn

and

2

i0

2 1

2ina

keEn


2

i0

2

2

f0

2 1

2

1

2

11

na

ke

na

ke

hc

2

i

2

f0

2 11

2

1

nna

ke

hc

2

i

2

f0

2 11

2 nnhca

keand HR

hca

ke

0

2

2

2

i

2

f

111

nnRH

(14)

where17 m 10097.1constant sRydberd': HR

nn of valuefinal: f

nn of valueinitial: i


Note:

For the hydrogen line spectrum,

Lyman series( nf=1 )

Balmer series( nf=2 )

Paschen series( nf=3 )

Brackett series( nf=4 )

Pfund series( nf=5 )

To calculate the shortest wavelength in any series, take ni= .

2

i

2

1

1

11

nRH

2

i

2

1

2

11

nRH

2

i

2

1

3

11

nRH

2

i

2

1

4

11

nRH

2

i

2

1

5

11

nRH


The Bohr‟s model of hydrogen atom

predicts successfully the energy levels of the hydrogen atom butfails to explain the energy levels of more complex atoms.

can explain the spectrum for hydrogen atom but some details ofthe spectrum cannot be explained especially when the atomis placed in a magnetic field.

cannot explain the Zeeman effect (Figure 11.7).

Zeeman effect is defined as the splitting of spectral lineswhen the radiating atoms are placed in a magnetic field.

5.3.3 Limitation of Bohr’s model of hydrogen atom

Magnetic

field

Transitio

ns

No magnetic

field

1

2Energy

Levels

Spectra

Figure 7


The Balmer series for the hydrogen atom corresponds to electronic transitionsthat terminate at energy level n=2 as shown in the 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)

Example 5 :

)eV(nE0.0

38.0

85.0

51.1

40.3

6

n

54

3

2

54.0

Figure 8


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

2

i

2

f

111

nnRH

22

7

max 3

1

2

110097.1

1

m 1056.6 7max

2f n


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

2

i

2

f

111

nnRH

22

7

min

1

2

110097.1

1

m 1065.3 7min

2f n

hc

EE if1


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 :

By applying the equation of wavelength for Lyman series, thus

Example 6 :

1; 3 fi nn

2

i

2

1

1

11

nRH

22

7

3

1

1

110097.1

m 1003.1 7


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

n

vn

16 s m 102.2

Technology

Phy 310 chapter 5