chapter 4 atomic structure - universitetet i...

“Atomic Structure”

Radio city broadcasts on a frequency of 5,090 KHz.What is the wavelength of electromagnetic radiation emitted by the transmitter?

Illustrative problem 1

cλ =

ν8

3

3 105090 10

×λ =

×20.589 10= ×

58.9 m=

s/m103isc 8×

The ratio of the energy of a photon of 2000 Å wavelength radiation to

that of 6000 Å radiation is

(a) ¼ (b) 4

(b) ½ (d) 3

Illustrative Problem 2

hcE h= ν =λ 1 2

1 2

hc hcE E= =λ λ

1 2

2 1

EE

λ∴ =

λ

0

0

6000 A 32000 A

= =

Hence, answer is (d).

Solution:

Illustrative Problem 3

The energy of the electron in the second and third Bohr orbits of the hydrogen atom is -5.42 X 10-12 and –2.41 X 10-12 respectively. Calculate the wavelength of the emitted radiation, when the electron drops from third to second orbit.

Solution 2 1E E E∆ = −

According to Planck’s quantum theory

12 125.42 10 ( 2.41 10 )ergs− −= − × − − ×123.01 10 ergs.−= − ×

E h= υ

cE h=λ

27h 6.6 10 ergs−= ×

Solution

56.6 10 cm−λ = ×

3 06.6 10 A= ×

0 81A 10 cm−=

27 8

126.62 10 3 10

3.01 10

−

−× × ×

=×

chE

∴ =λ

Class Exercise - 1 Which of the following fundamental particles are present in the nucleus of an atom? (a) Alpha particles and protons (b) Protons and neutrons (c) Protons and electrons (d) Electrons, protons and neutrons

Solution

The nucleus of an atom is positively charged and almost the entire mass of the atom is concentrated in it. Hence, it contains protons and neutrons.

Hence, answer is (b).

Class Exercise - 2 The mass of the proton is (a) 1.672 × 10–24 g (b) 1.672 × 10–25 g (c) 1.672 × 1025 g (d) 1.672 × 1026 g

Solution

Hence, answer is (a).

The mass of the proton is 1.672 × 10–24 g

Class Exercise - 3 Which of the following is not true in case of an electron? (a) It is a fundamental particle (b) It has wave nature (c) Its motion is affected by magnetic field (d) It emits energy while moving in orbits

Solution

Hence, answer is (d).

An electron does not emit energy while moving in orbit. This is so because if it would have done that it would have eventually fallen into the nucleus and the atom would have collapsed.

Class Exercise - 4 Positive charge of an atom is (a) concentrated in the nucleus (b) revolves around the nucleus (c) scattered all over the atom (d) None of these Solution

Hence, answer is (a).

Positive charge of an atom is present entirely in the nucleus.

Class Exercise - 5 Why only very few a-particles are deflected back on hitting a thin gold foil?

Solution

Due to the presence of a very small centre in which the entire mass is concentrated.

Class Exercise - 6 Explain why cathode rays are produced only when the pressure in the discharge tube is very low.

Solution

This is happened because at higher pressure no electric current flows through the tube as gases are poor conductor of electricity.

Class Exercise - 7 If a neutron is introduced into the nucleus of an atom, it would result in the change of (a) number of electrons (b) atomic number (c) atomic weight (d) chemical nature of the atom Solution

Hence, answer is (c).

Neutrons contribute in a major way to the weight of the nucleus, thus addition of neutron would result in increase in the atomic weight.

Class Exercise - 8 The concept of stationary orbits lies in the fact that (a) Electrons are stationary (b) No change in energy takes place in stationary orbit (c) Electrons gain kinetic energy (d) Energy goes on increasing

Solution

Hence, answer is (b).

When an electron revolves in a stationary orbit, no energy change takes place. Energy is emitted or absorbed only when the electron jumps from one stationary orbit to another.

Class Exercise - 9 What is the energy possessed by 1 mole of photons of radiations of frequency 10 × 1014 Hz?

Solution

E = hv E = 6.6 × 10–34 × 10 × 1014

E = 66 × 10–20 = 6.6 × 10–19 joules

∴ energy of 1 mole of photons = 6.6 × 10–19 × 6.023 × 1023

= 39.7518 × 104

= 397.518 kJ/mol

16

The Atomic Mass Scale and Mass Weights

amu) .9(0.691)(62 Mass atomicisotopeCu 63

=

17

The Atomic Weight Scale and Atomic Weights

isotopeCu isotopeCu 6563

amu) .9(0.309)(64 amu) .9(0.691)(62 weight atomic +=

18


copperfor amu 63.5 weight atomic

amu) .9(0.309)(64 amu) .9(0.691)(62 weight atomicisotopeCu isotopeCu 6563

=

+=

Atomic Number Atomic number (Z) of an element is the number of protons in the nucleus of each atom of that element.

Element # of protons Atomic # (Z)

Carbon 6 6

Phosphorus 15 15

Gold 79 79

Question 2 The table on the next slide shows the five isotopes of germanium found in nature, the abundance of each isotope, and the atomic mass of each isotope.

Calculate the atomic mass of germanium.

Answer 72.591 amu

The wavelength of the peak in the intensity graph is given by Wien’s law (T must be in kelvin):

Wien’s Displacement Law

EXAMPLE: Finding peak wavelengths

QUESTIONS:

EXAMPLE :Finding peak wavelengths

EXAMPLE : Finding peak wavelengths

EXAMPLE :Finding peak wavelengths

The Electron Volt • Consider an electron

accelerating (in a vacuum) from rest across a parallel plate capacitor with a 1.0 V potential difference.

• The electron’s kinetic energy when it reaches the positive plate is 1.60 x 10−19 J.

• Let us define a new unit of energy, called the electron volt, as 1 eV = 1.60 x 10−19 J.

Energy of an electron

QUESTION:

EXAMPLE 38.5 Energy of an electron

EXAMPLE : Energy of an electron

Example 1 Write out the electron configuration of the elements (i) carbon (Z = 6) (ii) magnesium (Z = 12) (iii) argon (Z = 18) Solution (i) 1s2 2s2 2p2 (ii) 1s2 2s2 2p6 3s2 (iii) 1s2 2s2 2p6 3s2 3p6 Example 2 Determine which of the following are in the excited state (i) 1s2 2s2 2p5 3s1 (ii) 1s2 2s2 2p6 3s2 3p6 4s1 (iii) 1s2 2s2 2p3 (iv) 1s2 2s2 2p6 3s1 3p5 Solution (i) and (iv) are in excited states

Mass Number Mass number is the number of protons and neutrons in the nucleus of an isotope: Mass # = p+ + n0

Nuclide p+ n0 e- Mass #

Oxygen -

10

- 33 42

- 31 15

8 8 18 18

Arsenic 75 33 75

Phosphorus 15 31 16

Information in the Periodic Table

The number at the bottom of each box is the average atomic mass of that element.

This number is the weighted average mass of all the naturally occurring isotopes of that element.

Question 1 How does the atomic number of an element differ from the element’s mass number?

Answer The atomic number of an element is the number of protons in the nucleus. The mass number is the sum of the number of protons and neutrons.


The atomic weight of an element is the weighted average of the masses of its stable isotopes

Example 5-2: Naturally occurring Cu consists of 2 isotopes. It is 69.1% 63Cu with a mass of 62.9 amu, and 30.9% 65Cu, which has a mass of 64.9 amu. Calculate the atomic weight of Cu to one decimal place.

Calculating Atomic Mass


Copper exists as a mixture of two isotopes.

The lighter isotope (Cu-63), with 29 protons and 34 neutrons, makes up 69.17% of copper atoms.

The heavier isotope (Cu-65), with 29 protons and 36 neutrons, constitutes the remaining 30.83% of copper atoms.


The atomic mass of Cu-63 is 62.930 amu, and the atomic mass of Cu-65 is 64.928 amu.

Use the data above to compute the atomic mass of copper.

Calculating Atomic Mass First, calculate the contribution of each isotope to the average atomic mass, being sure

to convert each percent to a fractional abundance.


The average atomic mass of the element is the sum of the mass contributions of each isotope.

Can you tell me: how many atoms are there in 43.2 grams of iron? Using unit analysis: The unit of the answer is atoms. The unit of the information given is grams. But there is no conversion factor for grams ---> atoms. Reason: the mass of an atom is different for every element. Solution: use the mole concept.

43.2g Fe x 1 mol Fe55.85 g Fe

x 6.02x1023 atoms Fe1 mol Fe

= 4.66x 1023 atoms Fe

Electron Orbits Consider the planetary model for electrons which move in a circle around the positive nucleus. The figure below is for the hydrogen atom.

Coulomb’s law:

2

204C

eFrπε

=

Centripetal FC:

2

2CmvFr

=

2 2

204

mv er rπε

= Radius of Hydrogen atom 2

204er

mvπε=

FC +

-

Nucleus

e- r

Example 1: Use the Balmer equation to find the wavelength of the first line (n = 3) in the Balmer

series. How can you find the energy?

2 2

1 1 1 ; 32

R nnλ

= + =

R = 1.097 x 107 m-1

2 2

1 1 1 1(0.361); 2 3 0.361

R RR

λλ

= + = =

7 -1

10.361(1.097 x 10 m )

λ = λ = 656 nm

The frequency and the energy are found from:

c = fλ and E = hf

Example 2: Find the radius of the Hydrogen atom in its most stable state (n = 1).

2 20

2nn hr

meε

π=

m = 9.1 x 10-31 kg

e = 1.6 x 10-19 C

2

22 -12 34 2Nm

C-31 -19 2

(1) (8.85 x 10 )(6.63 x 10 J s)(9.1 x 10 kg)(1.6 x 10 C)

rπ

− ⋅=

r = 5.31 x 10-11 m r = 53.1 pm

Total Energy of an Atom The total energy at level n is the sum of the kinetic and potential energies at that level.

221

20

; ; 4

eE K U K mv Urπε

= + = =

Substitution for v and r gives expression for total energy.

2

02nev

nhε=

2 20

2nn hr

meε

π=

But we recall that:

4

2 2 208nmeE

n hε= −

Total energy of Hydrogen atom for level n.

Energy for a Particular State It will be useful to simplify the energy formula for a particular state by substitution of constants.

m = 9.1 x 10-31 kg

e = 1.6 x 10-19 C

εo = 8.85 x 10--12 C2/Nm2

h = 6.63 x 10-34 J s

2

2

4 -31 -19 4

2 2 2 -12 2 2 -34 2C0 Nm

(9.1 x 10 kg)(1.6 x 10 C)8 8(8.85 x 10 ) (6.63 x 10 Js)n

meEn h nε

= − = −

-18

2

2.17 x 10 JnE

n= − 2

13.6 eVnE

n−

=Or

Balmer Revisited 4

2 2 208nmeE

n hε= −

Total energy of Hydrogen atom for level n.

Negative because outside energy to raise n level.

When an electron moves from an initial state ni to a final state nf, energy involved is:

4 4

0 2 2 2 2 2 20 0 0

1 1; 8 8f

f

hc me meE E Ehc h n h nλ λ ε ε

−= = − = +

4 4

2 3 2 2 2 2 3 20 0

1 1 1 ; If 8 8f f i f

me meRh cn n n h cnλ ε ε

= − =

Balmer’s Equation: 7 -1

2 20

1 1 1 ; 1.097 x 10 mf

R Rn nλ

= − =

Energy Levels We can now visualize the hydrogen atom with an electron at many possible energy levels.

Emission

Absorption

The energy of the atom increases on absorption (nf > ni) and de-creases on emission (nf < ni).

Energy of nth level: 2

13.6 eVEn

−=

The change in energy of the atom can be given in terms of initial ni and final nf levels:

2 20

1 113.6 eVf

En n

= − −

Spectral Series for an Atom The Lyman series is for transitions to n = 1 level.

The Balmer series is for transitions to n = 2 level.

The Pashen series is for transitions to n = 3 level.

The Brackett series is for transitions to n = 4 level.

n =2

n =6

n =1

n =3 n =4

n =5 2 20

1 113.6 eVf

En n

= − −

Example 3: What is the energy of an emitted photon if an electron drops from the n = 3 level

to the n = 1 level for the hydrogen atom?

2 20

1 113.6 eVf

En n

= − −

Change in energy of the atom.

2 2

1 113.6 eV 12.1 eV1 3

E = − − = −

∆E = -12.1 eV

The energy of the atom decreases by 12.1 eV as a photon of that energy is emitted.

You should show that 13.6 eV is required to move an electron from n = 1 to n = ∞.

Summary (Cont.)

Balmer’s Equation: 7 -1

2 20

1 1 1 ; 1.097 x 10 mf

R Rn nλ

= − =

653 nm 486 nm 410 nm

434 nm Spectrum for nf = 2 (Balmer)

n = 3 n = 4 n = 5 n6

The general equation for a change from one level to another:

Emission spectrum

Questions

a. Sketch a cathode-ray tube and label it with the following parts: partially evacuated glass vessel, electrically charged plates, cathode, anode, fluorescent screen, high voltage source.

b. Explain how it works.

1.

Questions

2. Do you think the hypothesis inferring from the observations are reasonable? Why or why not? Explain in each hypothesis

Questions

3. How can the observations from the cathode ray tube experiment lead to the new atomic model “the plum pudding model”

Plum Pudding Model • in 1904 J.J. Thomson proposed the plum pudding model of

atom.

• According to this model the electrons were randomly stuck into a ball of positively charge matter.

positively charge matter

electrons © 2004 by Pearson Education Inc.

• A more modern name for this model might be chocolate-chip muffin model.

57 Discovering the Electron

The Discovery of the Electron

Thomson found q/m for cathode ray particles to be 1.76 x 1011 C/kg This is close to 2000 x q/m for a hydrogen ion Thomson’s conclusion: cathode-ray particles (today called electrons) are approximately 1/2000 the mass of a hydrogen ion (today called protons)


Since Thomson didn’t know the charge or mass of these particles couldn’t use to determine their speed He balanced so that particle path was undeflected

= 212qV mv

versus e mF F


Then, Using only a magnetic field,

=

=

=

e mF F

q E qv B

Ev

B=

=

=

2

m cF F

vqv B m

rq vm Br


Examples: Practice Problem 1, Practice Problem 1,

446.0 10

2.4 102.50

NC m

s

Ev

TB×

= = = ×

681.0 10

1.0 101.0 0.0100

ms C

kgq vm Br T m

×= = = ×

×

6.3 Volts

120 Volts Plate is heated and electrons boil off. Velocity= 0 Potential Energy= ½ mv^2

Electrons are attracted to positively charged plate. They accelerate towards it and small percentage escape the plate through small hole, creating electron beam.

The potential energy of electrons is converted to kinetic energy

Since change in energy is the voltage times the charge then ½mv²=qV Therefore v= √(2qV/m)

√

We now know that v= √(2qV/m), so we can now easily find the velocity of our beam of electrons.

q(charge) of an electron= -1.6•10^-19

V(volts)=120

m(mass) of an electron=9.11•10^-31 kg

Therefore: v=√(2)(-1.6•10^-19)(120)/(9.11•10^-31)

v=√4.215•10^13

v=649•10^6 m/s

In order to predict the angle at which the electrons are deflected, we must first measure the force that the magnetic field inserts upon the beam

To do this, we use the equation: F=qvB

Magnetic field

Electrons

Like Solar Wind, the electrons in the CRT beam are deflected when entering a magnetic field, therefore the electron beam “bends.”

The force is always Perpendicular to the magnetic field And the velocity of the electrons

In order to find the force of the magnetic field, we must first calculate its strenghth.

mass= 9.11•10^-31 kg

velocity= 6.492•10^6 m/s

And if the measured distance of the electron beam from the magnets Is .075 meters

Then B= (9.11•10^-31)(6.492•10^6)/(1.6•10^-19)(.075)

B=2.772•10^-6 tesla

Since F=qvB and, according to Newton’s second law, F=m•v²/r, we can deduce that qvB=m•v²/r Or B=mv/qr

Charge= 1.6•10^-19 C

Now that we know the strength of the magnetic field at the electron beam, we can Calculate the force which the field exerts upon the electrons.

F=qvB

F=(6.49•10^6) (1.6•10^-19)(2.772•10^-6

F=2.879•10^-18 N

Examples 1. In a Cathode ray tube an electron travels through an electrical

field of 5.80 x 103 N/C and a magnetic field of 6.60 x 10-3 T undeflected.

a. what is the velocity of the electron?

b. If the electric field is turned off the magnetic field causes the electron to travel in a circular path, what is the radius of the circular path of the electron?

2. An electron travels through a CRT and a magnetic field of 4.20 x 10-3 T, the electron travels a circular path with a radius of 4.90 x 10-3 m. a. What is the velocity of the electron?

b. What was the potential difference in the CRT?

3. An unknown particle travels in a CRT tube through a magnetic field of 4.60 x 10-2 T at a velocity of 5.20 x 105 m/s, the magnetic field causes the particle to travel in a circular path with a radius of 0.235 m.

What is the charge to mass ratio of this particle?

Deflection of Electrons in a Uniform Electric Field (1)

Consider an electron beam directed between two oppositely charged parallel plates as shown below.

With a constant potential difference between the two deflecting plates, the trace is curved towards the positive plate. +

-

d


The force acting on each electron in the field is given by

deVeEF P==

where E = electric field strength, V = p.d. between plates, d = plate spacing.

p


The vertical displacement y is given by

22 )(21

21 t

mdeV

aty p==

2

2

)(21

vx

mdeVp=

This is the equation for a parabola.

Deflection of electrons in a uniform magnetic field

Deflection of Electrons in a Uniform Magnetic Field

The force F acting on an electron in a uniform magnetic field is given by

BevF =

Since the magnetic force F is at right angles to the velocity direction, the electron moves round a circular path.

Deflection of Electrons in a Uniform Magnetic Field

The centripetal acceleration of the electrons is

mBeva =

Hence m

Bevr

va ==2

which gives

eBmvr =

Determination of Specific Charge Using a Fine Beam Tube

and the kinetic energy of the electron provided by the electron gun is

eVmv =2

21

Where V is the anode voltage.

mBerv = (For an electron moving in a

uniform magnetic field) Since

Determination of Specific Charge Using a Fine Beam Tube (3)

Rearrange the equation gives

22

2rBV

me

=

The value of the specific charge of an electron is now known accurately to be

1110)000003.0758803.1( ×± C/kg

eVm

Berm =2)(21So

Thomson’s e/m Experiment (1)

Thomson’s apparatus for measuring the ratio e/m

× × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × ×

+

-

v

http://www-outreach.phy.cam.ac.uk/camphy/electron/electron_exp.htm�

Thomson’s e/m Experiment (2)

A beam of electron is produced by an electron gun with an accelerating voltage V.

The electron beam is arranged to travel through an electric field and a magnetic field which are perpendicular to each other.

The apparatus is set-up so that an electron from the gun is undeflected.

Thomson’s e/m experiment (3)

As the electron from the gun is undeflected, this gives

BE FF =

i.e. BeveE =BEv =⇒

On the other hand, eVmv =2

21

Combining the equations, we get 2

2

2VBE

me

=

Bev

eE

v

A B

d

+V

F +Q

position d

To move the charge from A to B a force equal to F would be needed. The work done moving Q from A to B is W.

W = Fd

But E = F/Q or F = EQ

So, W = EQd

Potential difference is defined as work done per unit charge:

V = W/Q

So: V = W/Q = EQd /Q V = Ed

E = V/d

V

(Uniform fields)

Acceleration: gravitational and electrical

A falling ball An accelerating charge

starts at highgravitationalpotential energy

gravitationalfield g

height h

gainskineticenergy

mass m

ends at lowgravitationalpotential energy

gravitationalpotential

ball falls down gravitationalpotential hill

gain of kinetic energy = loss of potential energy = mgh

electricpotential

charge ‘falls down’ electricalpotential hill

gain of kinetic energy = loss of potential energy = qV

starts at highelectricalpotentialenergy

potentialdifference V

gainskineticenergy

electricfield

ends at lowelectricalpotential energy

+

–

+

An electric field accelerates a charge as a gravitational field accelerates a mass

+charge q

ΔEp=mgΔh ΔW = QΔV

Draw the forces acting on the drop

Can you come up with an expression for the charge on the droplet?

E

Iden

tify

the

forc

es o

n a

char

ge in

an

elec

tric

field

Draw the forces acting on the drop

Can you come up with an expression for the charge on the droplet?

F= Eq

F= mg

q = mg E

Iden

tify

the

forc

es o

n a

char

ge in

an

elec

tric

field

1) electric force upward = weight downward EQ = mg Q = mg/E = 1.8 x 10-15 x 9.81 / 1.9 x 104 = 9.4 x 10-19 C

ΔEp=m gΔh

ΔW = Q ΔV

Quantization of Charge

Example: Practice Problem 1, page 763 (Pearson Physics) 2

5 14

19

19

19

5.0 10 2.4 10 9.81

4.7 10

4.7 103

1.60 10

e g

N mC s

Ce

F F

E m g

kg

q

q

q C

Ce

−

−

−

−−

−

=

=

× × = × ×

= ×

×=

×

Example

A beam of electrons travels an undeflected path in a cathode ray tube. E is 7.0 x 103 N/C. B is 3.5 x 10-2 T. What is the speed of the electrons as they travel through the tube?

What we know: E = 7.0 x 103 N/C B=3.5 x 10-2 T

Equation: v = E/B

Substitute: v = (7.0 x 103N/A s) / (3.5 x 10-2 N/A m)

Solve! v = 2.0 x 105 m/s

Example An electron of mass 9.11 x 10-31 kg moves with a speed of 2.0 x

105 m/s across a magnetic field. The magnetic induction is 8.0 x 10-4 T. What is the radius of the circular path followed by the electrons while in the field?

What we know: M = 9.11 x 10-31 kg B=8.0 x 10-4 T v=2.0 x 105 m/s

Equation: Bqv = mv2/r so r = (mv) / (Bq)

Substitute: R = (9.11x10-31kg)(2.0x105 m/s) (8.0x10-4N/Am)(1.6x10-19As)

Solve! r = 1.4 x 10-3 m

Modifications of the Bohr Theory – Elliptical Orbits

Sommerfeld extended the results to include elliptical orbits Retained the principle quantum number, n

Determines the energy of the allowed states Added the orbital quantum number, ℓ

ℓ ranges from 0 to n-1 in integer steps All states with the same principle quantum number are said

to form a shell The states with given values of n and ℓ are said to form a

subshell

Example

A negatively charged polystyrene sphere of mass 3.3 x 10-15 kg is held at rest between 2 parallel plates separated by 5.0 mm when the potentialdifference between them is 170 V. How many excess electrons are on the sphere?

Solution: Using qV/d = mg gives q = 9.5 x 10-19 C Taking e to be 1.6 x 10-19, q corresponds to 5.9 excess electrons Because we cannot have 0.9 of an electron, the answer must be 6 electrons, the discrepancy must be due to experimental inaccuray or rounding errors

Example

Electrons are accelerated to a speed of 8.4 x 106 m/s. they then pass into a region of uniform magnetic flux density 0.50 mT. The path of the electrons in the field is a circle with a radius 9.6 cm. Calculate:

a) the specific charge of the electron b) the mass of the electron, assuming the charge on the electron is 1.6 x

10-19 C Solution a) e/m = v/Br = 1.8 x 1011 C/kg b) using e/m, m = 9.1 x 10-31 kg

Example

It is required to select charged ions which have a speed of 4.2 x 106 m s-1. the electric field strength in the velocity selector is 3.2 x 104 V m-1. Calculate the magnetic field strength required.

Solution v = E/B = 7.6 x 10-3 T

Example

A charged particle has mass 6.7 x 10-27 kg and charge +3.2 x 10-19 C. It is travelling at speed 2.5 x 108 m s-1 when it enters a region of space where there is a uniform magnetic field of flux density 1.6 T at right angles to its direction of motion. Calculate

a) the force on the particle due to the magnetic field b) the radius of its orbit in the field

Solution a) F = Bqv sin θ = 1.3 x 10-10 N b) centripetal force is provided by the electromotive force mv2/r = Bqv r = 3.2 m

chapter 4 atomic structure - universitetet i...

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