· Web viewRadiation is emitted by the hydrogen atom when the electron “jumps” from a more energetic initial state to a less energetic state. The “jump” can’t be visualized

Chapter ONE Atomic Structure and Energy Levels First year class Sameer Abdul Kadhim

Chapter One Atomic Structure and Energy Levels

1-1 Atomic Model

In order to explain many phenomena associated with material we must know

the atom structure . Atomic models were proposed to explain the distributions of the

charged particles in an atom.

1-2 Early Models for Atom

The model of the atom in the days of Newton was a tiny, hard, indestructible

sphere. Although this model was a good basis for the kinetic theory of gases, new

models had to be devised when later experiments revealed the electronic nature of

atoms.

J. J. Thomson, in 1898,

proposed that an atom possesses a

spherical shape (radius

approximately 10–10 m) in which

the positive charge is uniformly

distributed. The electrons are

embedded into it in such a manner

as to give the most stable

electrostatic arrangement .Many

different names are given to this

model, for example, plum pudding, raisin pudding

or watermelon. This model can be visualized as a

pudding or watermelon of positive charge with

plums or seeds (electrons) embedded into it. An

important feature of this model is that the mass of

the atom is assumed to be uniformly distributed

1


over the atom. Although this model was able to explain the overall neutrality of the

atom, but was not consistent with the results of photoelectric effect , spectrum of

hydrogen atom , X-ray , …… .

In 1911 Ernest Rutherford and his students Hans Geiger and Ernest Marsden

performed a critical experiment showing that Thomson’s model couldn’t be correct.

In this experiment he found that: the atom consists of a nucleus of positive charge that

contains nearly all the mass of the atom. Surrounding this central nucleus negatively

charged electrons .

There are two basic difficulties with Rutherford’s planetary model. First, an

atom emits certain discrete characteristic frequencies of electromagnetic radiation and

no others; the Rutherford model is unable to explain this phenomenon. Second, the

electrons in Rutherford’s model undergo a centripetal acceleration. According to

Maxwell’s theory of electromagnetism, centripetally accelerated charges revolving

with frequency f should radiate electromagnetic

waves of the same frequency. Unfortunately,

this classical model leads to disaster when

applied to the atom. As the electron radiates

energy, the radius of its orbit steadily decreases

and its frequency of revolution increases. This

leads to an ever-increasing frequency of emitted

radiation and a rapid collapse of the atom as the electron spirals into the nucleus.

1-3 Bohr's Models for Hydrogen Atom

At the beginning of the 20th century, scientists were perplexed by the failure

of classical physics to explain the characteristics of spectra. Why did atoms of a given

element emit only certain lines? Further, why did the atoms absorb only those wave-

lengths that they emitted?

Neils Bohr (1913) was the first to explain quantitatively the general features of

Hydrogen atom structure and its spectrum. Bohr’s model for Hydrogen atom is based

on the following postulates:

2


1. The atom has a massive positively-charged nucleus.

2. The electron in the hydrogen atom can move around the nucleus in a circular

path of fixed radius and energy. These paths are called orbits, energy level or

energy states. These orbits are arranged concentrically around the nucleus.

3. The electrons revolve in these orbits, influence of the Coulomb force of

attraction (between the nucleus and electron) which it's balanced by the

centrifugal force for revolving electron.

4. An electron cannot revolve round the nucleus in any arbitrary orbit but in just

certain definite quantization

orbits. Only those orbits are

possible (or permitted) for

which the Hydrogen atom

doesn’t emit energy in the

form of electromagnetic

radiation. Hence, the total

energy of the atom remains

constant. Such orbits are also

known as stationary orbits

(also called Bohr's stationary

orbits). those orbits are possible (or permitted) for which the orbital angular

momentum of the electron is equal to an integral multiple of h/2 i.e.

Angular momentum of electron in the nth orbit is

m vn rn=nh2 π

where n is an integer (n= 1, 2, 3 etc. for the first, second and third

orbits respectively.

5. Let the different permitted orbits have energies of E1, E2, E3 etc. The electron

can be raised from n = 1 orbit to any other higher orbit if it is given proper

amount of energy equal to energy difference between these two orbits.

3


6. Radiation is emitted by the hydrogen

atom when the electron “jumps” from a

more energetic initial state to a less

energetic state. The “jump” can’t be

visualized or treated classically. In

particular, the frequency f of the

radiation emitted in the jump is related

to the change in the atom’s energy and

is independent of the frequency of the

electron’s orbital motion . The

frequency and wavelength of the

emitted radiation is given by:

f =Ei−E f

hλ= hc

Ei−E f

where Ei is the energy of the initial state, E f is the energy of the final state, h is

Planck’s constant, and Ei>E f .

1-4 De-Broglie's Hypothesis for Bohr's Stationary Orbits

For more than a decade following Bohr’s publication, no one was able to

explain why the angular momentum of the electron was restricted to these discrete

values. Finally, de Broglie (1924) gave a direct physical way of interpreting this

condition. In his doctoral dissertation in 1924, Louis de Broglie postulated that,

because photons have wave and particle characteristics, perhaps all forms of matter

have both properties . This was a highly revolutionary idea with no experimental

confirmation at that time. According to de Broglie, all moving particles, such as

4


electrons, atoms, molecules,….., having an associates wave called De-Broglie wave

with wavelength:

λd=h

mv

where v is the velocity of the particle. And the momentum

P=mv= hλd

De Broglie's idea was that, if light can have both a particle and a wave nature,

perhaps electrons can too! Perhaps the quantization of the angular momentum of an

electron in the hydrogen atom was due to the wave nature of the electron. An electron

with a linear momentum p=mv would have a wavelength λd =h/p. This is now called

the de Broglie wavelength . This relationship applies not only to photons and

electrons, but as far as we know, to all particles!

With a formula for the electron wavelength, de Broglie was able to construct a

simple model explaining the quantization of angular momentum in the hydrogen

atom. In de Broglie's model, one pictures an electron wave chasing itself around a

circle in the hydrogen atom. If the circumference of the circle, 2π r did not have an

exact integral number of wavelengths, then the wave, after going around many times,

would eventually cancel itself out as illustrated in Figure (1-A). But if the

circumference of the circle were an exact integral number of wavelengths as

illustrated in Figure (1-B), there would be no cancellation. This would therefore be

one of Bohr's allowed orbits.

5


Figure (1) de-Broglie wave length of electron in Hydrogen atom

Suppose (n) wavelengths fit around a particular circle of radius rn as shown in

Figure(2),then we have:

n λd=2 π rn

Using the de Broglie formula λd =h/mv for the electron wavelength, we get:

nhm vn

=2 π rn m vn rn=nh2 π

Figure(2) de-Broglie wavelength of electron in each orbit

6


Normal, Excited and Ionized Atom

When electron in its inner most orbit (n = 1), then the atom is said to be in its

normal (or unexcited) state. Generally, if atom possesses proper energy (by heating

or electrical voltage or ….)Then electron jump from unexcited orbit (n=1) to higher

(excited) permitted orbits having n = 2, 3, 4 etc. This atom called excited atom. When

the electron is completely removed from the atom (due to high energy), the atom

is said to be ionized.

1-5 Atomic Energy Level Diagram

Instead of drawing various electron orbits to the scale of their radii as

in Fig. (3 a), it is customary to draw horizontal lines to an energy scale as shown in

Fig. (3 b) and such a diagram is called energy level diagram of an atom. In this array

of energies, the higher energies are at the top while the lower energies are at the

bottom. The various electron jumps between allowed orbits now become vertical

arrows between energy levels.

Notes:

1-Usually energy in atom is measured in eV instate of Joule where:1 eV=1 . 6×10−19 Joules

2- The first level (n=1) called ground state or unexcited state.

3- The above states (n>1) called excited state where first excited state at n=2 and second exited state at n=3 and so on.

4- Ionized state at n=∞ where energy level =0eV.

Therefore ionized energy (absorbed energy that made atom ion) will be E∞−E 1=−E 1

7


Example 1-1 : The six lowest energy levels of Sodium vapor are:{ -15, -12.9, -

11.81, -11, -10.9 and -10.74eV}. When a photon of wavelength 3300Ao is observed, a

three other photons are emitted. Find wave length of these three emitted photons.

Solution

Energy of observed photon is Ephoton=hf =hc

λ

=6 .62×10−34×3×108

3300×10−10=6.02×10-19J=3.76eV

This energy observed by an electron in ground state (whose energy =-15eV) therefore

the new energy of electron =-15+3.76=-11.24eV. This new energy made electron

jump up to new position lie between third and fourth level, therefore the electron must

jump down to third level, that made the photon will emit the energy difference where:

ΔE1=(-11.24)-(-11.81)=0.57eV=9.12×10-20 J

Wave length of first photon is

8

E2=-12.9

E1=-15

E3=-11.81

E4=-11Enew=-11.24 ΔE1

ΔE2

ΔE3

First photon

Third photon

Secondphoton

Incident photon

Illustration for example 1-1


λ 1= hcΔE 1

=6 . 62×10−34×3×108

9 .12×10−20=2.177×10-6m

The second photon will emit when electron jump from third level to second level

where:

ΔE2=(-11.81)-( -12.9)=1.09 eV=1.744×10-19 J

Wave length of second photon is:

λ 2= hc

ΔE 2=6 . 62×10−34×3×108

1.744×10−19=1.138×10-6m

The third photon will emit when electron jump from second level to ground level

where:

ΔE3=( -12.9)-(-15)=2.1 eV=3.36×10-19 J

Wave length of third photon is

λ 3= hcΔE3

=6 .62×10−34×3×108

3. 36×10−19=5.91×10-7m

1-6 Hydrogen Atom Energy Level Diagram

9


The above postulates concerning Bohr's atomic model can be utilized to

calculate not only the radii of different electron orbits but also the velocity and orbital

energy possessed by different electrons. We will use Hydrogen atom as example to

illustrate the energy levels because it contain a single electron in its orbit.

1-6-1 Radius of Orbits

Now, the stability of the atom requires that the centrifugal force acting on the

revolving electron be balanced by the electrostatic pull exerted by the positively

charged nucleus on the electron.

Centrifugal force = pull force (Coulomb's force)

m v2

r= e2

4 π ϵ0 r2

Where:

ϵ 0: permittivity of free space and e: electron charge

Also, according to Bohr's postulates, mvr = nh/2 .

Now we will substitute v= nh

2 π mr in above equation we will obtain :

rn=( h2ϵ 0

πme2 )× n2

From this equation we can see that value of possible radii is discrete then the above

equation can be rewrite:

rn=r1× n2

Where rn is the radius of nth orbit and r1 is the radius of first orbital =0.52×10-10 m .

1-6-2 Velocity of Electron in the Orbits

10


In the same way by substituting equation of radius in the Bohr's equation we

will find:

vn=e2

2 h ϵ 0× 1

n=v1×( 1

n )Where vn is the velocity of revolving electron in the nth orbit and v1 is the velocity of

electron in the first orbital = 2 .1818×106 m /sec

1-6-3 Electron Energy for Each Orbit

There two type of energy for electron in Hydrogen atom:

1- Kinetic Energy (K.E=12

mv2

) since electron is a moving mass.

By substituting vnin the K.E equation we will have

K . E= me4

8 h2ε 02×

1n2

Joules

2- Potential Energy (P.E) since electron is a negative charge move in electrical

field of positive nucleus.

P .E= −e2

4 πε0 r by substituting rn in this equation we will obtain

P . E=−me4

4 h2ε02 × 1

n2 Joules

Therefore the total energy will be:

En=P. E+K . E=−m e4

8 h2 ϵ02 × 1

n2

11

En=−21 .7×10−19

n2 =E1

n2 Joule


Where En is the total energy of electron in the nth orbital or energy of the nth

level Energy can found in term of e.V:

Now we can found energy for each orbital

E1 is the energy of first orbit (ground or stable state)= =-13.6eV.

E2 is first excited level is =-3.4eV,

.

.

E∞=0eV

That made ionized energy for Hydrogen atom = E∞- E1=13.6eV

The energy level diagram of Hydrogen atom is shown in the Figure below .

12

En=−13.6

n2 eV


What does the negative electronic energy (En) for hydrogen atom mean?

The energy of the electron in a hydrogen atom has a negative sign for all possible

orbits . What does this negative sign convey? This negative sign means that the

energy needed to free electron from atom . As example in the hydrogen atom

electron in a ground state energy of has energy -13.6eV. This means that the electron

needs at least 13.6eV of kinetic energy in order to become free of the nucleus.

As the electron gets closer to the nucleus (as n decreases), En becomes larger

in absolute value and more and more negative. The most negative energy value is

given by n=1 which corresponds to the most stable orbit. We call this the ground

state.

1-7 Spectrum Lines of Hydrogen

The spectrum of radiation emitted by a substance that has absorbed energy is

called an emission spectrum. Atoms, molecules or ions that have absorbed radiation

are said to be “excited”. To produce an emission spectrum, energy is supplied to a

sample by heating it or irradiating it and the wavelength (or frequency) of the

13


radiation emitted, as the sample gives up the absorbed energy, is recorded as shown in

Figure below.

The light emitted by a sample of excited atoms (or any other element) can be

passed through a prism and separated into certain discrete wavelengths. Thus an

emission spectrum, which is a photographic recording of the separated wavelengths is

called as line emission spectrum.

Line emission spectra are of great interest in the study of electronic structure.

Each element has a unique line emission spectrum. The characteristic lines in

atomic spectra can be used in chemical analysis to identify unknown atoms in the

same way as finger prints are used to identify people.

For Hydrogen atom line

spectrum can be calculated by using

Bohr's atomic model . If the electron

jumps from initial energy state ni with

energy Ei into final state energy state n f

with energy E f (where ni>n f ), it emits

a photon of wavelength λ given by

λ= hcE i−E f

Wave number can define as reciprocal

of wavelength

1λ=

Ei−E f

hc

14


Where Ei=−m e4

8h2 ϵ 02 × 1

n i2 and E f=

−m e4

8h2ϵ 02 × 1

n f2 that made:

1λ= −m e4

8ch3 ϵ 02 ×( 1

ni2 −

1nf

2 )

¿ RH ( 1nf

2−1ni

2 ) Where RH Rydberg constant

RH =m e4

8 ch3 ϵ 02=1.1×107 m-1

The Swedish spectroscopes, Johannes

Rydberg, drive this expression of all

series of lines in the hydrogen

spectrum.

We can use this expression to

evaluate the wavelengths for the

various series in the hydrogen

spectrum as shown in Figure .

For example, in the Balmer

series (Johann Balmer,1884, Swiss mathematics teacher ), n f=2 and ni=2,3,4 , ….

that made emitted wavelength given by:

λ=364.56( n i2

ni2−4 )nm

The Balmer series of lines are the only lines in the hydrogen spectrum which

appear in the visible region of the electromagnetic spectrum. The observed lines have

15


wavelength 656.21nm (Red), 486.07nm (blue/green) , 434.01nm (blue/violet) and

410.12 nm (violet)

In the same way series of lines that correspond to n f=1,3,4,5 are known as

Lyman, Paschen, Bracket and Pfund series, respectively, as give in Table

Example1-2

An electron in Hydrogen atom has Kinetic energy of 2.416×10-19 J, drop to ground

state.

Calculate:

1- Radius and velocity of electron orbit before dropping into ground state.

2- Wavelength of the emitted radiation due to dropping.

3- The ionized energy of Hydrogen.

4- Total energy of electron at fourth excited level.

Solution

16


1-K.E n=

21 .7×10−19

n2 → n=

⌈√21 . 7×10−19

2 .416×10−19 ⌉= 3 electron at 3rd level

Radius is r3=0.529×10-10×(3)2 =4.761×10-10 m

And velocity is v3 =

2. 1818×106

3 =

2- At 3rd level electron energy is E3=

−13 .6 eV32

=-1.511eV

When electron drop to ground state its new energy will be -13.6eV

Then:

λ= hcE3−E1

= 6 . 626×10−34×3×108

(−1 .511−(−13 .6) )×1. 6×10−19=1.027×10-7m

3- Ionized energy =E∞−E1=−E1 =13.6 eV

4- Forth excited state → n=5 →E5=

−13 .6 eV52

×1.6×10−19

=

Example1-3

An electron in Hydrogen atom at ground state absorbs photon with energy of 2.04×10-

18J.

Calculate:

a- The orbital radius and electron energy at excited state.

b- How many spectral lines are emitted if electron made all possible transition in

its return to ground state?

c- The highest and lowest frequencies of these spectral lines.

Solution

17


a- The electron new energy will be

En =E1+E(photon)

En=-13.6eV+

2. 04×10−18

1.6×10−19=-0.85=

−13. 6 eVn2

→n=4, electron at 4th level

Radius is r4=0.529×10-10×(4)2=

b- There are 4 possible ways to jump from 4th state to ground state

First way (4) → (3 ) → (2 ) → (1 )

Second way (4) → (2 ) → (1 )

Third way (4) → (3 ) → (1 )

Fourth way (4) → (1 )

Therefore there are 6 different possible spectra lines can be emitted.

b- Frequency of emitted radiation (

ΔEh ) depend on energy difference between

these level

Maximum radiation frequency =ΔE (max )

h =

E4−E1

h =3.078×1015 Hertz

Minimum radiation frequency =ΔE (min )

h =

E4−E3

h =1.593×1014 Hertz

Tutorial QuestionQ\1: Explain the main weakness points in nuclear atomic model

proposed by Rutherford and how could Bohr able to solve them.

18


Q\2: Explain the meaning Bohr's stationary orbits, and how could de

Broglie able to explain why the angular momentum of these orbits

should be discrete values.

Q\3: Explain De-Broglie's Hypothesis for Bohr's Stationary Orbits.

Q\4: Explain the meaning of negative electron energy for Hydrogen

atom and why first level is the most stable level.

Q\5: The six lowest energy levels of Sodium vapor are [-15, -12.9, -11.81, -11, -10.9 and -10.74eV]. When a photon of observed, a two other photons are emitted with wavelength (1.138 and 0.591 )µm, Find :1. Wavelength of observed photon.2. Find velocity of electron in of 3rd exited level If potential energy is

-22eV.3. Lowest and highest frequencies could be emitted when electron in 3 rd

excited state drop into ground state.

Q\6: Cold Mercury vapor is bombarded with radiation and as a result

the fluorescent line 2.537 and 4.078A0 appear. What is the wavelength

have been present in the bombarding photon ?

Q\7: An electron in Hydrogen atom with De-Broglie wavelength=9.690A0 drop into ground state , Find:

1- The orbital radius and electron energy before dropping into ground state.

2- How many spectral lines are emitted if electron made all possible transition in its return to ground state?

3- The highest and lowest frequencies of these spectral lines.4- Ionized energy of Hydrogen atom.

Q\8: An electron in Hydrogen atom has centrifugal force 1.0109×10-9

N, find:

19


1- Number of revolution /sec for the electron

2- The longest photon wavelength would be required to ionize atom.

3- Number of spectral lines if electron made all possible transition in

its return to ground state?

Q\9: For an electron revolving in the Hydrogen atom, find number of

revolution /sec for first three orbits. (Note: number of revolution per

second = velocity of electron in this orbital /circumference of orbit)

Q\10: What is the wavelength of photon that able to ionize a Hydrogen

atom in the normal state and give the ejected electron a kinetic energy of

3.4 eV.

Q\11: Prove that wavelength of spectrum lines in the Balmer series are

given by:

Then calculate the wave-number for the longest emitted wavelength in the Balmer series.Q\12: Prove that for Hydrogen atom kinetic energy for any energy level is given by :

Q\13: Prove that De-Broglie wavelength for electron revolving in

Hydrogen atom is given by :

20


Q\14: Prove that the time for one revolution of electron in the Hydrogen atom is:

T= √me2

4 √2 ε° (−En )1.5

Where Enis the total energy ( in Joule) of the nth level.

21

Documents

· Web viewRadiation is emitted by the hydrogen atom when the electron “jumps” from a more energetic initial state to a less energetic state. The “jump” can’t be visualized