PERUZI NASI-TEACHING AND LEARNING NOTES: PHYSIC FORM 6 ATOMIC PHYSICS- STRUCTURE OF THE ATOM

PHYSIC FORM 6 ATOMIC PHYSICS- STRUCTURE OF THE ATOM

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Monday, August 15, 2022

PHYSIC FORM 6 ATOMIC PHYSICS- STRUCTURE OF THE ATOM

UNAWEZA JIPATIA NOTES ZETU KWA KUCHANGIA KIASI KIDOGO KABISA:PIGA SIMU: 0787237719

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ATOMIC PHYSICS- STRUCTURE OF THE ATOM

THOMSON’S MODEL OF ATOM

According to Thomson an atom is a positive charged sphere in which the entire mass and positive charge of the atom is uniform distributed with negative electrons embedded in it as shown.

$C:\thlb\cr\tz\ATOMIC PHYSICS_files\image001.jpg$

The number of electrons is such that their negative charge is equal to the positive charge of the atom. This atom is electrically neutral.

This model was called Thomson’s plum pudding model because the negatively charge electrons (the plums) were embedded in a sphere of uniform positive charge (the pudding).

Drawbacks of this Model
1.It could not provide stability to the atom it is because the positive and negative charges are stationary and will be drawn towards each other, thus destroying the individual negative and positive charges.
2. It could not explain the presence of discrete spectral lines emitted by hydrogen and other atoms.

RUTHER FORD’S MODEL OF ATOM
The salient features of this model are
(i)Every atom consist of a tiny central core, called the nucleus which contains all the atom’s positive charge and most of its mass (99.9%).
$C:\thlb\cr\tz\__i__images__i__\at1.png$ $C:\thlb\cr\tz\__i__images__i__\at2.png$

(iii) The electrons occupy the space outside the nucleus. Since an atom is electrically neutral the positive charge on the nucleus is equal to the negative charge on electrons surrounding the nucleus.

(iv) Electrons are not stationary but revolve around the nucleus in various circular orbits as do the planets around the sun.

In this way Rutherford provided stability to the atom. It is because the centripetal force required by the electrons for revolution is provided by the electrostatic force of attraction between electrons and the nucleus.

$C:\thlb\cr\tz\__i__images__i__\18_7.jpg$

e = charge on electron

z =total number of protons in the nucleus

m=mass of the electron

r =distance of electron from the nucleus

v= linear velocity of the electron

Force of attraction between electron and the nucleus is

$C:\thlb\cr\tz\ATOMIC PHYSICS_files\image005.gif$

$C:\thlb\cr\tz\ATOMIC PHYSICS_files\image006.gif$

where Ze is a nuclear charge

The centripetal force required to keep the electron moving in circular path is

$C:\thlb\cr\tz\__i__images__i__\at3.png$

Since the atom is stable $C:\thlb\cr\tz\ATOMIC PHYSICS_files\image009.gif$

$C:\thlb\cr\tz\__i__images__i__\at4.png$
$C:\thlb\cr\tz\__i__images__i__\at5.png$

Kinetic energy of electron

$C:\thlb\cr\tz\__i__images__i__\at6.png$

From equation (1)

$C:\thlb\cr\tz\__i__images__i__\4454.PNG$

$C:\thlb\cr\tz\ATOMIC PHYSICS_files\image014.gif$
$C:\thlb\cr\tz\__i__images__i__\at8.png$
$C:\thlb\cr\tz\__i__images__i__\at9.png$

Potential energy of electron

$C:\thlb\cr\tz\__i__images__i__\at11.png$

Total energy of electron

$C:\thlb\cr\tz\__i__images__i__\at12.png$

$C:\thlb\cr\tz\ATOMIC PHYSICS_files\image019.gif$

$C:\thlb\cr\tz\__i__images__i__\at13.png$

The total energy of electron in the orbit is negative hence the electron is bound to the positive nucleus

For hydrogen Atom

For hydrogen atom z= 1. Therefore K. E and P.E OF electron in hydrogen atom are

$C:\thlb\cr\tz\__i__images__i__\at14.png$

The total energy of electrons hydrogen atom is

$C:\thlb\cr\tz\ATOMIC PHYSICS_files\image023.gif$

$C:\thlb\cr\tz\ATOMIC PHYSICS_files\image024.gif$
$C:\thlb\cr\tz\__i__images__i__\at15.png$

Limitations of Rutherford’s model of atom

1. According to Maxwell’s theory of electromagnetism a charge that is accelerating radiates energy as electromagnetic waves
The electron moving around the nucleus is under constant accelerating radiates energy as electromagnetic waves.

– Due to this continuous loss of energy the electrons in Rutherford’s model were bound to spiral towards the nucleus and fall into it when all of their rotational energy were radiated

– Hence Rutherford’s atomic model cannot be stable while in actual practice, an atom is stable

This shows that Rutherford’s model is not correct

1. During inward spiraling the electron’s angular frequency continuously increases

– As result electrons will radiate electromagnetic waves of all frequency i.e. the spectrum of these waves will be continuous in nature because these are continuous loss of energy.

– But this is contrary to observation experiments shows that an atom emits line spectra and each line corresponds to a particular frequency or wavelength.

Rutherford’s model failed to account for the stability of the atom. It was also unable to explain the emission of line spectra.

BOHR’S MODEL OF ATOM
According to Bohr’s atomic model, the revolving electrons in the atom do not emit radiations under all conditions. They do so under certain conditions as expalined by him in his model.

BASIC POSTULATES OF BOHR’S MODEL OF ATOM

1. The electrons revolve around the nucleus of the atom in circular orbits. The centripetal force required by electrons for revolution is provided by the electrostatic force of attraction between the electrons and the nucleus.

2. An electron can revolve only in those circular orbits in which its angular momentum is an integral multiple of $C:\thlb\cr\tz\ATOMIC PHYSICS_files\image026.gif$

$C:\thlb\cr\tz\__i__images__i__\at16.png$

h= Plank’s constant.

Radius of orbit r

From,
$C:\thlb\cr\tz\__i__images__i__\at16.png$

$C:\thlb\cr\tz\__i__images__i__\at17.png$

Since n is a whole number only certain value of r is allowed.

Thus according to Bohr, an electron can revolve only in certain orbits of definite radii not in all these are called stable orbits (stationary orbit)

According to this postulate the angular momentum of the electron does not have continuous range i.e. the angular momentum of the revolving electron is quantized.

While revolving in stable or stationary orbits the electrons do not radiate energy inspite of their acceleration towards the centre of the orbit.

– For this reason these permitted orbits are called stable or stationary orbits.

e= charge on electron

m= mass of electron

r_n= radius of the n^th orbit

v_n= velocity of electron in the n^th orbit

Z= number of positive charge (protons)

Positive charge on nucleus Ze

RADIUS OF BOHR’S STATIONARY ORBITS

As the centripetal force is provided by the electrostatic force of attraction between the nucleus and electron.

$C:\thlb\cr\tz\__i__images__i__\at18.png$

$C:\thlb\cr\tz\__i__images__i__\at19.png$

According to Bohr

$C:\thlb\cr\tz\__i__images__i__\at20.png$

Consider equation
$C:\thlb\cr\tz\__i__images__i__\at21.png$

Take equation (ii) square it

$C:\thlb\cr\tz\__i__images__i__\at20.png$

$C:\thlb\cr\tz\__i__images__i__\at22.png$

Take equation (iii) $C:\thlb\cr\tz\ATOMIC PHYSICS_files\image039.gif$ equation (i)

$C:\thlb\cr\tz\ATOMIC PHYSICS_files\image040.gif$ = $C:\thlb\cr\tz\ATOMIC PHYSICS_files\image041.gif$ . $C:\thlb\cr\tz\ATOMIC PHYSICS_files\image042.gif$

$C:\thlb\cr\tz\__i__images__i__\at23.png$

It is clear that $C:\thlb\cr\tz\ATOMIC PHYSICS_files\image045.gif$ $C:\thlb\cr\tz\ATOMIC PHYSICS_files\image046.gif$ n², radii of the stationary orbits are in ratio 1²: 2²:3²………..clearly the stationary orbits are not equally spaced.

For hydrogen atom

For hydrogen atom z= 1, so that equation become
$C:\thlb\cr\tz\__i__images__i__\at24.png$

Now $C:\thlb\cr\tz\ATOMIC PHYSICS_files\image048.gif$ = 0.53 x10^-10m

$C:\thlb\cr\tz\__i__images__i__\at25.png$

e = electronic charge $C:\thlb\cr\tz\ATOMIC PHYSICS_files\image045.gif$ = (0. 53 x 10^-10) $C:\thlb\cr\tz\ATOMIC PHYSICS_files\image049.gif$ metres

$C:\thlb\cr\tz\__i__images__i__\at26.png$

Thus the radii of the first, second and third stationary orbits of hydrogen atom are 0.53 Å, 2.12 Å and 4.77Å respectively.

2. VELOCITY OF ELECTRON IN BOHR’S STATIONARY ORBIT

From equation below, we have

$C:\thlb\cr\tz\__i__images__i__\at27.png$

$C:\thlb\cr\tz\__i__images__i__\at28.png$

Putting the value of $C:\thlb\cr\tz\ATOMIC PHYSICS_files\image045.gif$ into that equation

$C:\thlb\cr\tz\__i__images__i__\at29.png$

$C:\thlb\cr\tz\__i__images__i__\at30.png$

It is clear that $C:\thlb\cr\tz\__i__images__i__\at31.png$ in other words, electrons move at a lower speed in higher orbits and vice versa.

For hydrogen atom

Z =1

Then
$C:\thlb\cr\tz\__i__images__i__\at32.png$

3. FREQUENCY OF ELECTRON IN STATIONARY ORBIT

The number of revolution completed per second by the electron in a stationary orbit around the nucleus

Velocity of electron in the $C:\thlb\cr\tz\__i__images__i__\4455.PNG$ orbit

$C:\thlb\cr\tz\__i__images__i__\at33.png$

For hydrogen atom

Z = 1

Then,
$C:\thlb\cr\tz\__i__images__i__\at34.png$

Frequency of electron in the first orbit of hydrogen atom is n=1, r₁=0.53×10^-10_m

$C:\thlb\cr\tz\__i__images__i__\at35.png$

Electron in first orbit of hydrogen atom will have a frequency of 6.57x 10¹⁵revolutions per second.

4. TOTAL ENERGY OF ELECTRON IN STATIONARY ORBIT

The total energy E_nof the electron in the n^th orbit is the sum of kinetic and potential energy in the n^th orbit.

– The K.E of electron in the n^th orbit is

$C:\thlb\cr\tz\__i__images__i__\142.png$

The potential energy of electron in the n^th orbit is

$C:\thlb\cr\tz\__i__images__i__\4456.PNG$

Total energy of electron in the n^th orbit is

$C:\thlb\cr\tz\__i__images__i__\317.png$
$C:\thlb\cr\tz\__i__images__i__\410.png$

But

$C:\thlb\cr\tz\__i__images__i__\58.png$

$C:\thlb\cr\tz\__i__images__i__\66.png$

$C:\thlb\cr\tz\__i__images__i__\74.png$

Thus as n increases i.e. electron moves to higher orbit, the total energy of the electron increases i.e. total energy becomes less negative.

For hydrogen atom z=1

$C:\thlb\cr\tz\__i__images__i__\82.png$

Thus the total energy of electron in a stationary orbit is negative which means that the electron is bound to the nucleus and it is not free to leave the atom.

We can find the total energy of electron in the various orbits of hydrogen atoms as under.

$C:\thlb\cr\tz\__i__images__i__\4457.PNG$

The total energy of electron increases i.e. becomes less negative as the electron goes to higher orbits

When n→∞ E_n =0 and the electron becomes free

Ground state/ normal state

This is the state of atom when the entire electrons in it occupies their lowest energy levels as required by their n and l values.

The energy of an atom is least i.e. largest negative value when n=1 i.e. when electron revolves in the first orbit.

The energy of hydrogen atom in the ground state is 13.6eV.

Excited state

This is the state of an atom when electrons in an atom occupy energy levels higher than those permitted by the values of n and l values.

At room temperature most of the hydrogen atoms are in the ground state

If hydrogen atom absorbs energy i.e. due to rise in temperature it may be promoted to one of the higher orbits (i.e. n=2, 3, 4…..)

The atom is said to be in the excited state.

WAVE LENGTHOF EMITTEDRADIATION.

When an electron jumps from a higher orbit (n₂) to the lower orbit (n₁) the energy difference between the two orbits is released because the energy of electron in the higher orbit is more than in the lower orbit. Consider two orbits having principle quantum numbers n₂and n₁ where n₂>n₁

Then energy of electron in the two orbits is given by

$C:\thlb\cr\tz\__i__images__i__\1110.png$

As the electron jumps from orbit n₂ to n_1,energy is released in the form of electromagnetic radiation.

$C:\thlb\cr\tz\__i__images__i__\1214.png$

where

f= frequency of the emitted radiation

$C:\thlb\cr\tz\__i__images__i__\143.png$
The wavelength of the emitted radiation is given by

c=λf

$C:\thlb\cr\tz\ATOMIC PHYSICS_files\image110.gif$ = $C:\thlb\cr\tz\ATOMIC PHYSICS_files\image111.gif$

$C:\thlb\cr\tz\__i__images__i__\151.png$

This equation gives the wavelength of emitted radiation.

Now,

$C:\thlb\cr\tz\ATOMIC PHYSICS_files\image110.gif$ = $C:\thlb\cr\tz\ATOMIC PHYSICS_files\image113.gif$ = wave number

$C:\thlb\cr\tz\__i__images__i__\161.png$

Wave number

These are the number of waves in a unit length.

For hydrogen atom

For hydrogen atom z = 1

$C:\thlb\cr\tz\__i__images__i__\171.png$

This gives the mathematical formula for the wavelength of radiation emitted by hydrogen atom when electron jumps from outer orbit to inner orbit.

$C:\thlb\cr\tz\__i__images__i__\181.png$

where

R_His Rydberg constant. The value of R_H can be calculated as the value of e, m, h and c are known

$C:\thlb\cr\tz\__i__images__i__\191.png$

HOW TO CALCULATE THE RYDBERG CONSTANT USING CALCULATOR

From

$C:\thlb\cr\tz\__i__images__i__\201.png$
$C:\thlb\cr\tz\__i__images__i__\2111.png$

Clearly, wavelength/frequency of radiation emitted from the excited atom is not continuous. They have definite value depending upon the values of $C:\thlb\cr\tz\ATOMIC PHYSICS_files\image128.gif$ , and $C:\thlb\cr\tz\ATOMIC PHYSICS_files\image129.gif$

SPECTRAL SERIES OF HYDROGEN ATOM

Bohr gave a mathematical explanation for the spectrum of hydrogen atom.

The whole hydrogen spectrum can be divided into district groups of lines each group of lines is called spectral series.

The wavelength of the lines in each group can be calculated from Bohr’s formula

$C:\thlb\cr\tz\ATOMIC PHYSICS_files\image110.gif$ = $C:\thlb\cr\tz\ATOMIC PHYSICS_files\image130.gif$ $C:\thlb\cr\tz\ATOMIC PHYSICS_files\image112.gif$

The following are spectral series of hydrogen atom

i) Lyman series

ii) Balmer series

iii) Paschen series

iv) Bracket series

v) Pfund series

i) Lyman series

The Lyman series is obtained when electron jump to first orbit n₁=1 from outer orbits ( $C:\thlb\cr\tz\ATOMIC PHYSICS_files\image129.gif$ =2, 3, 4…)

Therefore the formula for calculating the wavelength of the lines in this series is,

$C:\thlb\cr\tz\__i__images__i__\229.png$

where

$C:\thlb\cr\tz\__i__images__i__\231.png$

This series lies in the ultraviolet region which is the invisible region.

ii) Balmer series

$C:\thlb\cr\tz\__i__images__i__\241.png$ $C:\thlb\cr\tz\__i__images__i__\252.png$

Therefore the formula for calculating the wavelength of the lines in this series is $C:\thlb\cr\tz\__i__images__i__\254.png$

$C:\thlb\cr\tz\__i__images__i__\261.png$

where

$C:\thlb\cr\tz\__i__images__i__\271.png$

This series lies in the visible spectrum and was found first of all in the hydrogen series

iii) Paschen series

$C:\thlb\cr\tz\__i__images__i__\281.png$ $C:\thlb\cr\tz\__i__images__i__\291.png$

Therefore the formula for calculating the wavelength of the lines in this series is
$C:\thlb\cr\tz\__i__images__i__\30.png$

where

$C:\thlb\cr\tz\__i__images__i__\4458.PNG$

This series lies in the infrared region.

iv) Brackett series

$C:\thlb\cr\tz\__i__images__i__\331.png$ $C:\thlb\cr\tz\__i__images__i__\341.png$

Therefore the formula for calculated the wavelength of the lines in this series is

$C:\thlb\cr\tz\__i__images__i__\351.png$

This series lies in the infrared region.

v) Pfund series
The Pfunds series is obtained when electrons jump to fifth orbit n₁ = 5 from outer orbits (n₂=6, 7, 8…..)

Therefore the formula for calculating the wavelength of the lines in this series is
$C:\thlb\cr\tz\__i__images__i__\371.png$

where

(n₂=6, 7, 8…..)

This series also lies in the infrared region

ENERGY LEVEL DIAGRAM

Energy level diagram is a diagram in which the total energies of electron in different stationary orbit of an atom represented by parallel horizontal lines drawn according to some suitable energy scale

In order to draw energy level diagram of an atom we must know the total energy of electron in different stationary orbits.

The total energy of an electron in the n^th orbit of hydrogen atom is given by

$C:\thlb\cr\tz\__i__images__i__\381.png$

By putting value of n=1, 2, 3….. we can find the total energy of electron in various stationary orbits of hydrogen atom as

$C:\thlb\cr\tz\__i__images__i__\391.png$

Similarly we can find the total energy of electron in the higher orbits

The table below gives the total energy of electron of hydrogen atom in different stationary orbits.
$C:\thlb\cr\tz\__i__images__i__\40.png$

$C:\thlb\cr\tz\__i__images__i__\4459.PNG$

The energy level diagram of hydrogen atom is shown below

Total energy of electron in a stationary orbit is represented by a horizontal line drawn to some suitable energy scale.

(i) The hydrogen atom has only one electron and this normally occupies the lowest level and has energy of -13.6eV

When the electron is in this level the atom is said to be in the ground state.At room temperature nearly all the atoms of hydrogen are in ground.

(ii) If hydrogen atom absorbs energy (due to rise in temperature )the electron may be promoted into one of the higher energy levels

The atom is now said to be in an excited state.Thus when the electron occupies other than the lowest energy level the atom is said to be in the excited state.

(iii) Once in an excited state the atom is unstable after a short time interval the electron falls back into the lowest state so that the atom is again in the ground state.

The energy that was originally impacted is emitted as electromagnetic waves.

(iv) The total energy of electron for (n= $C:\thlb\cr\tz\ATOMIC PHYSICS_files\image149.gif$ ) it becomes free of atom.

The minimum energy required to free the electron from the ground state of an atom is called ionization energy

For hydrogen atom ionization energy is +13. 6eV

(v) The difference between the adjacent energy goes on decreasing as the value of n increases.

So much so that when n>10 the energy difference is almost zero this is show by closeness of energy level lines at higher levels.

(vi) Note that region is labeled continuous at energy above zero n= $C:\thlb\cr\tz\ATOMIC PHYSICS_files\image149.gif$ level, the electron is free from the atom and is at rest

Higher energy represents the translation kinetic energy of the free electron

This energy is not quantized and so all energies above n = $C:\thlb\cr\tz\ATOMIC PHYSICS_files\image149.gif$ are allowed

IMPORTANT TERMS

It is desirable to discuss some important terms much used in the study of structure of atom.

(i) EXCITATION ENERGY

Excitation energy is the minimum energy required to excite an atom in the ground state to one of the higher stationary state.

Hydrogen atoms are usually in their lowest energy state where n=1

In this state (ground state) they are said to be unexcited.

However if you bombard the atoms with particles such as electron or proto collision can excite them

In other words a collision may give an atom enough energy to change it from ground state to some higher stationary state.Consider the case of hydrogen atom we know that $C:\thlb\cr\tz\ATOMIC PHYSICS_files\image151.gif$ = -13.6eV (ground state $C:\thlb\cr\tz\ATOMIC PHYSICS_files\image152.gif$ = -3.4eV (first excited state) $C:\thlb\cr\tz\ATOMIC PHYSICS_files\image153.gif$ =1.51eV (second excited state) and $C:\thlb\cr\tz\ATOMIC PHYSICS_files\image154.gif$ =0

In order to lift an electron from ground state n =1 to the first excited state n=2 energy required is E

E = $C:\thlb\cr\tz\ATOMIC PHYSICS_files\image155.gif$ – $C:\thlb\cr\tz\ATOMIC PHYSICS_files\image156.gif$

E= -3.4 – (- 13.6)

E = 10.2eV

Therefore the bombarding particle must provide an energy of 10.2eV to excite the atom from n =1 state to n=2 state

Similarly to excite the atom from n=1 state to n=3 state energy required is

$C:\thlb\cr\tz\__i__images__i__\411.png$

E = – 1.51 – (-13.6)

E = 12.1eV

We say that first and second excitation energies of hydrogen are 10.2eV and 12.1eV respectively

(ii) EXCITATION POTENTIAL

Excitation potential is the minimum accelerating potential which provide an electron energy sufficient to jump from the ground state n=1 to one of the outer orbits

$C:\thlb\cr\tz\__i__images__i__\4460.PNG$
Energy required to lift a electron from ground state n=1 to n=2 state is
$C:\thlb\cr\tz\__i__images__i__\441.png$
$C:\thlb\cr\tz\__i__images__i__\451.png$

Hence excitation potential for the first excited state of hydrogen is 10. 2V

Similarly energy required to lift an electron from ground state n=1 to n=2

$C:\thlb\cr\tz\__i__images__i__\461.png$

The value of excitation potential depend upon the state to which the atom is excited to which the atom is excited from the ground state

(iii) IONIZATION ENERGY

Ionization energy is the minimum energy needed to ionized an atom

Consider the case of hydrogen atom it has only one electron and this normally occupies the ground state.

The energy of the electron for n= $C:\thlb\cr\tz\ATOMIC PHYSICS_files\image149.gif$ state is zero and if the electron is lifted to this level (n= $C:\thlb\cr\tz\ATOMIC PHYSICS_files\image149.gif$ ) it becomes free of hydrogen atom i.e. hydrogen atom is ionized

$C:\thlb\cr\tz\__i__images__i__\4461.PNG$

(iv) IONIZATION POTENTIAL

Ionization potential is the minimum accelerating potential which would provide electron energy sufficient to just remove the electron from the atom.

$C:\thlb\cr\tz\__i__images__i__\481.png$

The ionization potential of one electron atom or ion is given by

$C:\thlb\cr\tz\__i__images__i__\491.png$

(v) QUANTIZATION OF ENERGY

Quantization of energy is the existence of energy radiated by atoms in a specific amount which is are integral multiples of a constant (hf).

SUCCESS OF BOHR’S THEORY

The success of bohr’s theory is not to be attributed so much to the mechanical picture of atom he proposed but rather to the development of mathematical explanation that agrees exactly with experimental observations. Bohr’s theory achieved the following successes.

i) MADE ATOM STABLE

Bohr’s theory made the atom stable according to this theory an electron moving in the formatted (quantum) orbits cannot lose energy even though under constant acceleration. This provided stability to the atom.

ii) INTRODUCED QUANTUM MECHANICS

Bohr’s theory introduced quantum mechanics in the realm of atom for the first time

Bohr’s explained that sub- atomic particles e.g. electrons are governed by the laws of quantum mechanics and not by classical laws of electron hydrogen as assumed by Rutherford

This completely changed our thinking and was the major step towards the discovery of the rudiment laws of the atomic world

iii) GAVE MATHEMATICAL EXPLANATION OF HYDROGEN SERIES

The hydrogen series found by various scientists were based on empirical relation but had no mathematical explanation

However these relations were easy derived by applying Bohr Theory

Further the size of hydrogen atom as calculated from this theory agreed very closely with the experimental value.

LIMITATIONS OF BOHR’S THEORY

Bohr’s simple theory of circular orbits inspire of its many successes was found inadequate to explain many phenomena observed experimentally.

This theory suffered from the following drawbacks.

(i) It could not explain the difference in the intensities of emitted radiations.

(ii) It is silent about the wave properties of electron

(iii) It could not explain experimentally observed phenomena such as Zeeman Effect, Stack effect etc.

(iv) Bohr’s model does not explain why the orbit are circular while elliptical path is also possible

(v) It could only partially explain hydrogen atom. For example this theory does not explain the fine structure of spectral lines in the hydrogen atom

PHYSIC FORM 6 ATOMIC PHYSICS- STRUCTURE OF THE ATOM

WORKED EXAMPLES

1. 1. Find the radius of the first orbit of hydrogen atom. What will be the velocity of electron in the first orbit? Hence find the size of hydrogen atom

Solution

The radius of n^th orbit of it atom is given by

$C:\thlb\cr\tz\__i__images__i__\50.png$

Radius of first orbit of it atom n=1

$C:\thlb\cr\tz\__i__images__i__\511.png$

Velocity of electron in the n^thorbit of hydrogen atom is given by

$C:\thlb\cr\tz\ATOMIC PHYSICS_files\image180.gif$ = $C:\thlb\cr\tz\ATOMIC PHYSICS_files\image181.gif$

Velocity of electron in the first orbit of hydrogen atom is given by

$C:\thlb\cr\tz\__i__images__i__\521.png$

Since there is one electron in hydrogen atom the size hydrogen atom is equal to double the radius of the first orbit

Size of the atom

= 2 $C:\thlb\cr\tz\ATOMIC PHYSICS_files\image050.gif$

= 2 x 0.53Å

Size of an atom =1.06Å

2. (a) The hydrogen atom is stable in the ground, state why?

(b) The ionization energy of hydrogen is 13. 6eV what does it mean?

Solution

If the hydrogen atom is in the ground state (n=1) there is no state of lower energy to which a down ward transition can occur thus a hydrogen atom in the ground state is stable

a) It means that energy required to remove the single electron from the lowest energy state of hydrogen atom to becomes free electron is 13.6eV

b) Second line of Lyman series is obtained when electron jumps from third orbit $C:\thlb\cr\tz\ATOMIC PHYSICS_files\image129.gif$ =3 to the first orbit n=1

According to Bohr’s theory the wavelength of emitted radiation is given by

$C:\thlb\cr\tz\ATOMIC PHYSICS_files\image186.gif$ = $C:\thlb\cr\tz\ATOMIC PHYSICS_files\image187.gif$ $C:\thlb\cr\tz\ATOMIC PHYSICS_files\image188.gif$

$C:\thlb\cr\tz\ATOMIC PHYSICS_files\image189.gif$ = $C:\thlb\cr\tz\ATOMIC PHYSICS_files\image190.gif$ $C:\thlb\cr\tz\ATOMIC PHYSICS_files\image191.gif$

$C:\thlb\cr\tz\ATOMIC PHYSICS_files\image192.gif$ = $C:\thlb\cr\tz\ATOMIC PHYSICS_files\image190.gif$ x $C:\thlb\cr\tz\ATOMIC PHYSICS_files\image193.gif$

$C:\thlb\cr\tz\__i__images__i__\531.png$

3.( a) What is the meaning of negative energy of orbiting electron?

(b) What would happen if the electron in atom were stationary?

$C:\thlb\cr\tz\__i__images__i__\541.png$

Solution
a)   The negative total energy means that it is bound to the nucleus. If it acquires enough energy from some external source (a collision for example) to make its total energy zero the electron is no longer bound it is free.
b)     If the electrons were stationary they would fall into the nucleus due to electrostatic force of attraction so atom would be unstable i.e. it would not exist
c)     For Paschen series we have longest wavelength line $C:\thlb\cr\tz\ATOMIC PHYSICS_files\image129.gif$ =4

This is a wavelength in the infrared part. Other lines in this series have shorter wavelength bad approach series limit of wavelength to given by This wavelength is also in the hydrogen part. This the range or centre series (820.4nm to 1875nm) is the infrared

4. a) If an electron jumps from first orbit to third orbit will it absorb energy?
b) Name the series of hydrogen spectrum lying in the infrared region
c) Calculate the shortest wavelength of the Balmer series
d) What is the energy possessed by an electron for n=?

Solution

a) Yes it is because the energy level of third orbit is more than that of the first orbit

b) * Paschen series

* Bracket series

* P fund series

Solution
In Balmer series the radiation of shortest wavelength (i.e. of highest of highest energy) is emitted when electron jumps from infinity orbit $C:\thlb\cr\tz\ATOMIC PHYSICS_files\image129.gif$ = $C:\thlb\cr\tz\ATOMIC PHYSICS_files\image149.gif$ to the second orbit $C:\thlb\cr\tz\__i__images__i__\552.png$ =2 of hydrogen atom.

5. a) The ionization potential of hydrogen is 13.6V what does it mean?

b) Find the longest wavelength in Lyman series

c) How much is the ionization potential of hydrogen atom?

d) The energy of the hydrogen atom in the ground state is 13.6eV. Determine the energies of those energy levels whose quantum numbers are 2 and 3.

Solution

a) The ionization energy of hydrogen is 13.6eV. Therefore, if an electron which has been accelerated from rest through a p.d of 13.6V collides with a hydrogen atom it has exactly the right amount of energy to produce ionization.

This is a common method of producing ionization and therefore the term ionization potential is often used.

b) Solution

In Lyman series the radiation of longest wavelength (i.e. lowest energy) is emitted when electron jumps from second orbit $C:\thlb\cr\tz\ATOMIC PHYSICS_files\image129.gif$ =2 to first orbit n=1 of hydrogen atom

c) The energy of hydrogen atom in the ground state is – 13.6eV. therefore its ionization energy is 13.6eV and ionization potential =13.6V

d) Solution

The energy of an electron in the n^th orbit of hydrogen atom is given by

$C:\thlb\cr\tz\__i__images__i__\562.png$

6. a) Name the series of hydrogen spectrum lying in the

i) Visible region

ii) Utraviolet region of electromagnetic spectrum

b) Write the empirical relation for Paschen series lines of hydrogen spectrum

c) What are the values of first and second excitation potential of hydrogen atom?

d) Calculate the radii and the energy of three lowest energy allowed orbits for the electron in Lithium ion. What is the energy of a photon that when absorbed causes an electron in Lithium ion to be excited from n=1 to n=3 state?

Solution

a) i) Balmer series

ii) Lyman series

b) The wavelength of the spectral lines in paschen series are given by

c) Excitation energy for first excited state = -3.4 – (-13.6)

=10.2eV

For second excited state

= – (1. 51 – (-13.6)

= 12.1eV

Solution

d) (I ) for a single electron atom or ion the radius of the n^th orbit is given

$C:\thlb\cr\tz\__i__images__i__\581.png$

$C:\thlb\cr\tz\__i__images__i__\59.png$

For a single electron atom or ion the energy of electron in the nth orbit is given by

$C:\thlb\cr\tz\__i__images__i__\60.png$

Thus the energy of n=1, 2 and 3 orbits. The photo energy must be equal to the energy needed to excite the electron

E₃ – E₁ =hf

(-13.6) – (-122.4) =photon’s energy

Photon’s energy = 108.8eV

7. The ionization energy of hydrogen like atom is 4rydbergs

(a) What is the wavelength of radiation emitted when electron jumps from first excited state to the ground state?

(b)What is the radius of the first orbit for this atom?
$C:\thlb\cr\tz\__i__images__i__\611.png$
$C:\thlb\cr\tz\__i__images__i__\623.png$

(d) According to Bohr’s theory what is the angular momentum of a electron in the third orbit

Solution

The energy electron in the n^th orbit of hydrogen like atom is

$C:\thlb\cr\tz\__i__images__i__\631.png$

The energy required to excite the electron from n=1 level to n=2

$C:\thlb\cr\tz\__i__images__i__\641.png$

If is the wavelength of the emitted radiations then, Radius of first orbit for this atom

$C:\thlb\cr\tz\__i__images__i__\651.png$

Solution

( b) Radius of n^th orbit

$C:\thlb\cr\tz\__i__images__i__\4462.PNG$
$C:\thlb\cr\tz\__i__images__i__\67.png$

Angular momentum L of an electron in n^th orbit is

L = n $C:\thlb\cr\tz\ATOMIC PHYSICS_files\image250.gif$

Here n=3

Then

L= 3 $C:\thlb\cr\tz\ATOMIC PHYSICS_files\image251.gif$

L= $C:\thlb\cr\tz\ATOMIC PHYSICS_files\image252.gif$

8. The energy levels of an atom are shown in figure below.

$C:\thlb\cr\tz\__i__images__i__\4463.PNG$

9. (a)Which one of these transitions will result in the emission of photon of wavelength 275nm?

(b) An electron orbiting in hydrogen atom has energy level of 3.4eV what will be its angular momentum
(c) The total energy of an electron in the first excited state of hydrogen atom is about
– 3. 4eV what is the wavelength?
solution
(a) Energy of emitted photon E

$C:\thlb\cr\tz\__i__images__i__\731.png$

Therefore photon of wavelength 275nm will be emitted for transition B

Solution
$C:\thlb\cr\tz\__i__images__i__\741.png$

$C:\thlb\cr\tz\__i__images__i__\751.png$

(d) K.E of electron = -(total energy of electron)

K.E of electron =3.4eV

ii) P.E of electron = 2xtotal energy

P.E of electron = -6.8eV

10. (a) How many lines can be drawn the energy level diagram of hydrogen atom?

(b) Use Bohr’s model to determine the ionization energy of the He ion also calculate the minimum wavelength a photo must have to cause ionization

$C:\thlb\cr\tz\__i__images__i__\761.png$ $C:\thlb\cr\tz\__i__images__i__\77.png$

(d) i) In neon atom the energies of the 3s and 3p states are respectively 16.70eV and 18.70eV. What wavelength corresponds to 3p -3s transitions in neon atom?

ii) The wavelength of the first member of the Balmer series in hydrogen spectrum is 6563Å. Calculate the wavelength of first member of Lyman series in the same spectrum