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TOPIC 1: THE ATOM > ATOMIC SPECTRUM – CHEMISTRY FORM 5

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TOPIC 1: THE ATOM > ATOMIC SPECTRUM - CHEMISTRY FORM 5

TOPIC 1: THE ATOM > ATOMIC SPECTRUM – CHEMISTRY FORM 5

ATOMIC SPECTRUM: Definition: Are light waves which have definite line and colour because t

ATOMIC SPECTRUM

Definition: Are light waves which have definite line and colour because they have intermediate wave length which cannot be detected by a human eye.

These spectrums have no harmful affect on human. Atomic spectrum produced once the atom gain energy which cause the electron to be excited and jump from lowest energy level.

This results into an atom being unstable in order to maintain the stability these electrons return back to its group state which accompanied with release of energy inform of radiation. These radiation have wavelength which detected by a human eye of definite colour definite wave and definite line.

Continuous Spectrum: are spectrums which contain all possible frequency over wide range of energy. Continuous spectrums are colourless and have no definite line because they contain very short wave lengths which are not detected by an eye. The continuous spectrum recorded in a spectrographic plate as given below.

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Line spectrum: is a spectrum which consists of scattered definite line. These spectrums have very long wave length. The spectrographic plate of line spectrum

Band Spectrum: Band spectrum are spectrums which consists of a group of definite line in small bands. The spectrographic plate of band spectrum includes the following:

BOHR-ATOMIC THEORY

Word Image 69Word Image 70(i) H – SPECTRUM
Definition:

Is a definite line and colour which resulted when electric discharge is passed through hydrogen gas in the emission tube under very low pressure. The H – spectrum recorded in the spectrographic plate. The following is a horizontal diagram of H-spectrum

 

UV Violet Infrared 

Red

Red
X – ray Radio electron
    – ray Television 

wave

UV – ray
Invisible

Note: wave length increases

The explanation of horizontal diagram in terms of atomic structure:-

First band: is a colourless band or invisible band. These are spectrum produced by electrons which excited from the first shell. The electron from the first shell experience stronger nuclear attractive force which use high energy in order to jump toward highest energy level. When these electron returned back release high amount of energy which have shorter wave length. These wave lengths are not detected by a human eye. The radiations formed are continuous spectrum such as X – ray, sun rays etc.

Second band: is a visible band which has definite line and colour. These are line spectrum, produced by electron which excited from the second shell. The electron in the second shell need moderate energy in order to jump towards highest energy level when returned  back release a normal energy which its wave length detected by a human eye.

Third band: Is invisible band. These spectrums are colourless and have no definite line.The electrons which produce these spectrum caused the scattered spectrum which appear as a colourless band. This is due to the lowest energy and highest wave length. The following include vertical diagram of H-spectrum.

The vertical diagram of Hydrogen Spectrum

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Lyman series: is a series of spectrum which resulted by electron excited from the first shell (n =1). These series correspond to the invisible band or colourless band.

Balmer series: is a series of spectrum which resulted by the electron which excited from the second shell (n = 2) this corresponds with visible or violet band.

Paschen’s series: is a series of spectrum which resulted with electron excited from the third shell (n = 3) etc.

Bracket series: is a series of spectrum which resulted with electron which excited from the forth shell (n = 4) etc.

p-fund series
: is a series of spectrum which resulted with the electron which exited from the fifth shell (n=5).

PLANCK’S QUANTUM THEORY

Planck put forward the Planck’s quantum theory. This theory has three main points which include the following;-

i.   Any radiation should be association with energy.

ii.  The energy is released inform of radiation, occur in small packets called quanta.

iii.The energy is directly proportion to the frequency.

E  f

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E = hf……………………..(1)

Since          h = Planck’s constant

h = 6.63 x 10 -34 JS

n= 1

Since          f =

Word Image 221

E =  ………………….(2)

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C = 3.0 x 108 ms-1

REYDBERG EQUATION

Reydberg put forward a principle which applied to find wave length of spectrum. The wave length of H-spectrum determined is applied to find frequency and energy of the H- spectrum. The wave number is inversely proportional to the square of energy level differences (n2).

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V

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1/V = RH n2

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V =  RH

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Since   V = 1/λ
=  ………………. (3)

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Where λ= wave length
n= lowest energy level

n= highest energy level

RH = Reydberg constant RH is 1.09 x 107m-1

The value of n1 and nfor H-spectrum can be obtained if given number of line and number of series. The value n1 is equal number of series given. But the value n2 is equal to the number of line plus the number of series. Example the value of n1 and nof third line of Balmer’s series include the following:-

Number of line = Third line

Number of series = Balmer’s series

n1 = number of series = 2

n2 = number of line + Number of series

= 3 + 2

n = 5

But:

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= 1.09 X 107m-1[1/22-1/52]

= 1.09 X 107m-1[1/4-1/25]

= 1.09 X 107m-1[(25-4)/100]

= 1.09 X 107m-1[(21/100]
λ = 4.45 x 10-7m of third line

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E=

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E=6.3 x 10-3 Js x 3.0 x 108ms-1

4.45 x 10-7m

E=4.25 x 10-19J

Also;

f =

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f =

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f = 6.74 x 1014s-1

f = 6.74 x 1014Hz

TRANSITION ENERGY

Is the energy which is required to shift electron from one shell to another. The energy required to shift electrons from one shell to another should be equal to the energy difference between those two shells. The energy difference between lowest energy level E1 and highest energy level E2 is obtained by using the following expression;-

E = E2 – E1……………… (1)

The trend of transition of electrons include the following;

E = E2 – E1 – Transition take place from E1 – E2 exactly.

E > E2 – E1 – Transition take place from E1 toward above E2

E < E2 – E1 – Transition take place from E1 and hang between E1 and E2

But if electrons gain enough energy which is equal to the ionization energy result the electron to jump completely from ground state to infinite. These electrons cannot return back instead leave atom ionized positively.

The energy supplied to the atom results to be excited and electron move completely from the ground state to the infinite. The energy supplied to the electrons is used as ionization energy and as kinetic energy. The ionization energy of electrons is equal to the amount of energy associated by electron in the shell or quantum number where it belongs.

The following expression  is used to find the energy associated by electrons or energy of that shell.

1/λ = RH

Let n1 = n

n= ∞

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since 1/ = 0

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1/λ =   ………………………..(i)
E = hc/λ
E/hc = 1/λ ……………………..(ii)

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Substitute (ii) into (i)

=

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E =

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E =

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Since R, h and C are constant

1eV = 1.6 x 10-19J

1MeV = 106eV

1MeV = 1.6 x 10-13J

But discovered that the energy increase from the nucleus atom toward the high energy level. The energy increases from first shell toward seventh shell. From seventh shell the energy is zero or the energy at infinite is zero. Then the energy from infinite which is zero towards the lowest shell decreases and results to be negative value of energy

n∞       =            0.00ev

n7         =          -0.15ev

n6         =          -0.21ev

n5         =          -0.54ev

n4         =          -0.84ev

n3         =          -1.50ev

n2         =          -3.40ev

n1         =          -13.60ev

E = E2 – E1

= E – En

= 0 – En

= 0 –

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E =

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The negative value occurs because the energy of infinite is zero. Since energy increase from nuclear towards highest energy level result the energy of each shell be negative value. This formula is used to determine energy of electron at each shell.

Example

i.   Find the first ionization energy of potassium.

ii.  Find the wavelength, frequency and energy of third line of Balmer’s series ( RH = 1.09 x 107m-1)

iii.If the wavelength of the first number of Balmer’s series is 6563Å. Calculate Reydberg constant and wavelength of the first member of the Lyman series.

iv.Find energy associated with electrons in the quantum number 2.

Solution:

i.              E = I.E

k = 2:8:8:1

E =

Word Image 261

E =

Word Image 262

= -1.36 x 10-19J

E = E2 – E1

E = 0 – -1.36 x 10-19 J

E = 1.39 x 10-19J

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ii.    n1= 2

n2 = 3 + 1

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λ =

λ= 4.86 x 10-7m

iii.    λ = 6563A
From

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RH =

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RH = 1.09 x 10-3 A-1

RH = 1.9 x 10 -3 x 1010m-1

RH = 1.09 x 107m-1       

iv.      E =?

n = 2

E =

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E =  J

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E= -5.44 x 10-19J

Example

The U.V light has a wave length 2950 Å. Calculate its frequency energy 1A = 10-10M

Given  J

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J

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If the electron dropped from E4 to E2: Find the frequency and wave length energy release

Given line spectrum

Word Image 71

Wave length

i.  Which line has highest frequency and energy?

ii.  Which line has lowest frequency and energy?

iii.   Find the energy and frequency of each line in (i) and (ii) above?

E5
E4
E3
e0 E2
E1

State either or not the transition of electrons would occur if energy supplied is

i.   Greater than E4 – E2

ii.   Equal to E4 – E2

iii.  Less than E4 – E2

iv.  Greater than E3 – E2 but less than E4 – E2

v.   Smaller than E4 – E2

Solution:

a)       = 2950A

Word Image 283 Word Image 284

f = c

λ

f =

Word Image 285

f = 1.02×10-2 S-1

E= hf

E = 6.3 x 10-34Js x 1.02 x 10-2 S-1

E = 6.426 x 10-36J

b)      E4 = -1.36 x 10-19J

E2 = -15.44 x 10-19J

E = E2 – E1

E = E4 – E2

E = (-1.36 x 10-19) – (-5.44 x 10-19)

E = 4.08 x 10-1

But:

E = hf

f =

Word Image 286

f =

Word Image 287

f = 6.15 x 1014 S-1

Hence:

F =

Word Image 288

λ =

Word Image 289

λ =

Word Image 290

λ = 4.88 x 10-7m

C. i)     A line of wave length 2030A

ii)    A line of wave length 8092 A

iii)   f =?

E =?

Line of 2030A

Since               f =

Word Image 294

f =

Word Image 295

f = 1.48 x 1015S-1

E = hf

E = 6.3 x 10-34Js x 1.48 x 1015

E=9.32 X 10-19   J

D)  Solution

Word Image 296

i) Greater than E4 – Eresult transition of electron from E1 and above E4

ii)  Equal E4 – E2 transition take place from E2 to E4

iii)  Smaller than E4 – E2 result transition of electrons from E2 but cannot reach instead hang between E2 and E4

iv)  Greater than E3 – E2 but smaller than E– E2 result transition of electron from E2 and above E3 but cannot reach E4

THE QUANTUM THEORY

WAVE PARTICLE DUALITY NATURE OF MATTER

States that: “Matter has particle nature as well as wave nature”. This means that matter have dual properties or two properties such as particle nature and wave nature. The wave particle duality nature of matter was put forward by De Broglie scientist. De Broglie’s derived an expression which applied to find the De Broglie’s wave length.

De Broglie’s wave length is in terms of mass and momentum.

Hence from Einstein equation;

E = mc2 …………………….(i)

From Planck’s equation

E =  ……………… (ii)

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Compare equation (i) and (ii)

mc2 =

Word Image 298

λ  = 

Word Image 299

 

λ  = 

Word Image 300

De Broglie’s wave length in terms of energy

From:

λ =  multiply by 1/c in both side

Word Image 301

λ  =  [1/c]

Word Image 302 Word Image 303

=

Word Image 304 Word Image 305

 = 

Word Image 306Word Image 307

Since E = mc2

Example

Alpha particles emitted from radium have energy of 4.4MeV. What is the de-Broglie’s wave length?

The mass of moving particles is 9.01 x 10-19g. What is the de-Broglie’s wave length?

The momentum of particles is 2.0 x 10-10gm/s. What is the de-Broglie’s wave length?

Given that:

h = 6.63 x 10-34 JS

C = 3.0 x 108m/s

Solution

a.   E = 4.4MeV

λ =?

De Broglie’s

=

Word Image 308 Word Image 309

λ =

Word Image 310

λ = 4.52 x 10-13m

b.  Solution

m = 9.01 x 10-19g

λ =?

De Broglie’s

λ =

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λ =

Word Image 312

= 2.33 x 10-24m

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c.   P = 2.0 x 10-10gms

λ =? De Broglie’s

λ =

Word Image 314

λ =

Word Image 315

λ = 3.315 x 10-24Jg-1m-1s-1

HEISENBERG UNCERTAINITY PRINCIPLE

Heisenberg uncertainity principle state that, “It is impossible to determine position and momentum of electrons simultaneously with greater accuracy.” It is impossible to determined position and momentum of electrons because;-

(i)The size of electron is very small and as such radiations of high energy extremely small wavelength are required to detect it.

(ii) Impact of these high energy photons changes both the direction and speed of the electron.

Thus; the very act of measurement disturbs the position of electron. The uncertainties in the determination of these two quantities vary inversely. If one is determined fairly
accurately, the other must be corresponding less accurate.

The distance of electron: Is a position of electron from the nucleus and momentum of electrons is product of mass of electrons and velocity of the electron, Heisenberg put forward an expression which is used to determine uncertainity position and momentum of electrons.

Where: P – uncertainity momentum

X – Uncertainity position

P

Word Image 316 Word Image 317

P = x

Word Image 318 Word Image 319

P =  …………………. (1)

Word Image 320

Since= proportionality constant

Word Image 321

Also

mc = …………………. (2)

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Example

The uncertainity in the momentum of particles is 3.3 x 10-16gms-1. Find accuracy with which its position can be determined

X =

Word Image 323

X =

Word Image 324

= 1.59 x 10-29m

TRANSITION ENERGY

Is the energy which is required to shift electron from one shell to another. The energy required to shift electrons from one shell to another should be equal to the energy difference between those two shells. The energy difference between lowest energy level E1 and highest energy level E2 is obtained by using the following expression;-

E = E2 – E1……………… (1)

The trend of transition of electrons include the following;

E = E2 – E1 – Transition take place from E1 – E2 exactly.

E > E2 – E1 – Transition take place from E1 toward above E2

E < E2 – E1 – Transition take place from E1 and hang between E1 and E2

But if electrons gain enough energy which is equal to the ionization energy result the electron to jump completely from ground state to infinite. These electrons cannot return back instead leave atom ionized positively.

The energy supplied to the atom results to be excited and electron move completely from the ground state to the infinite. The energy supplied to the electrons is used as ionization energy and as kinetic energy. The ionization energy of electrons is equal to the amount of energy associated by electron in the shell or quantum number where it belongs.

The following expression  is used to find the energy associated by electrons or energy of that shell.

1/λ = RH

Let n1 = n

n= ∞

Word Image 162

since 1/ = 0

Word Image 163Word Image 164

1/λ =   ………………………..(i)
E = hc/λ
E/hc = 1/λ ……………………..(ii)

Word Image 165

Substitute (ii) into (i)

=

Word Image 166 Word Image 167

E =

Word Image 168

E =

Word Image 169

Since R, h and C are constant

1eV = 1.6 x 10-19J

1MeV = 106eV

1MeV = 1.6 x 10-13J

But discovered that the energy increase from the nucleus atom toward the high energy level. The energy increases from first shell toward seventh shell. From seventh shell the energy is zero or the energy at infinite is zero. Then the energy from infinite which is zero towards the lowest shell decreases and results to be negative value of energy

n∞       =            0.00ev

n7         =          -0.15ev

n6         =          -0.21ev

n5         =          -0.54ev

n4         =          -0.84ev

n3         =          -1.50ev

n2         =          -3.40ev

n1         =          -13.60ev

E = E2 – E1

= E – En

= 0 – En

= 0 –

Word Image 170

E =

Word Image 171

The negative value occurs because the energy of infinite is zero. Since energy increase from nuclear towards highest energy level result the energy of each shell be negative value. This formula is used to determine energy of electron at each shell.

Example

i.   Find the first ionization energy of potassium.

ii.  Find the wavelength, frequency and energy of third line of Balmer’s series ( RH = 1.09 x 107m-1)

iii.If the wavelength of the first number of Balmer’s series is 6563Å. Calculate Reydberg constant and wavelength of the first member of the Lyman series.

iv.Find energy associated with electrons in the quantum number 2.

Solution:

i.              E = I.E

k = 2:8:8:1

E =

Word Image 172

E =

Word Image 173

= -1.36 x 10-19J

E = E2 – E1

E = 0 – -1.36 x 10-19 J

E = 1.39 x 10-19J

Word Image 174

ii.    n1= 2

n2 = 3 + 1

Word Image 175

Word Image 176

Word Image 177Word Image 178

Word Image 179

λ =

λ= 4.86 x 10-7m

iii.    λ = 6563A
From

Word Image 180Word Image 181
Word Image 183

Word Image 184

Word Image 185Word Image 186

Word Image 187

RH =

Word Image 188

RH = 1.09 x 10-3 A-1

RH = 1.9 x 10 -3 x 1010m-1

RH = 1.09 x 107m-1       

iv.      E =?

n = 2

E =

Word Image 189

E =  J

Word Image 190

E= -5.44 x 10-19J

Example

The U.V light has a wave length 2950 Å. Calculate its frequency energy 1A = 10-10M

Given  J

Word Image 192

J

Word Image 193

If the electron dropped from E4 to E2: Find the frequency and wave length energy release

Given line spectrum

Word Image 67

Wave length

i.  Which line has highest frequency and energy?

ii.  Which line has lowest frequency and energy?

iii.   Find the energy and frequency of each line in (i) and (ii) above?

E5
E4
E3
e0 E2
E1

State either or not the transition of electrons would occur if energy supplied is

i.   Greater than E4 – E2

ii.   Equal to E4 – E2

iii.  Less than E4 – E2

iv.  Greater than E3 – E2 but less than E4 – E2

v.   Smaller than E4 – E2

Solution:

a)       = 2950A

Word Image 194 Word Image 195

f = c

λ

f =

Word Image 196

f = 1.02×10-2 S-1

E= hf

E = 6.3 x 10-34Js x 1.02 x 10-2 S-1

E = 6.426 x 10-36J

b)      E4 = -1.36 x 10-19J

E2 = -15.44 x 10-19J

E = E2 – E1

E = E4 – E2

E = (-1.36 x 10-19) – (-5.44 x 10-19)

E = 4.08 x 10-1

But:

E = hf

f =

Word Image 197

f =

Word Image 198

f = 6.15 x 1014 S-1

Hence:

F =

Word Image 199

λ =

Word Image 200

λ =

Word Image 201

λ = 4.88 x 10-7m

C. i)     A line of wave length 2030AWord Image 202

ii)    A line of wave length 8092 AWord Image 203

iii)   f =?

E =?

Line of 2030A

Since               f =

Word Image 205

f =

Word Image 206

f = 1.48 x 1015S-1

E = hf

E = 6.3 x 10-34Js x 1.48 x 1015

E=9.32 X 10-19   J

D)  Solution

Word Image 207

i) Greater than E4 – Eresult transition of electron from E1 and above E4

ii)  Equal E4 – E2 transition take place from E2 to E4

iii)  Smaller than E4 – E2 result transition of electrons from E2 but cannot reach instead hang between E2 and E4

iv)  Greater than E3 – E2 but smaller than E– E2 result transition of electron from E2 and above E3 but cannot reach E4

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