8-ch7(struktur atom).ppt

8/22/2019 8-Ch7(struktur atom).ppt

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ATOMIC STRUCTURE

All Bold Numbered Problems


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Chapter 7 Outline

Events leading to Quantum Mechanics

Newton

Planck

Einstein

Bohr

de Broglie

Schrdinger

Heisenberg

Using Quantum Numbers


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ATOMIC STRUCTURE

From the ERA of

Newtonian Physics to

Quantum Physics


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ELECTROMAGNETICRADIATION


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Electromagnetic Radiation

Most subatomic particles behave asPARTICLES and obey the physics ofwaves.

Define properties of waves

Figure 7.1 and7.2.

Wavelength, Node

Amplitude


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Figures 7.1

Electromagnetic Frequency


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wavelength Visible light

Wavelength ()Ultraviolet radiation

Amplitude

Node

There are

noLIMITS

to ...there are

an .


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Node in a

standing wave

wavelengt

hVisible light

wavelengt

hUltaviolet radiation

Amplitude

Nod

e


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Waves have a frequency

Use the Greek letter nu, , for frequency,and units are cycles per sec

All radiation: = cwhere c = velocity of light = 3.00x108 m/sec

Long wavelength ----> small frequency

Short wavelength ----> high frequency


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increasing

wavelength

increasing

frequency


Long wavelength -----> small frequency

Short wavelength -----> high frequency

See Figure 7.3


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Figure 7.3

Long wavelength -----> small frequency

Short wavelength -----> high frequency


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Red light has = 700. nm.Calculate the frequency.

Freq =3.00 x 10

8m/s

7.00 x 10-7

m= 4.29 x 1014 sec-1

700nm 1 x 10-9 m

1 nm= 7.00 x 10-7 m

Examples


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StandingWaves

See Figure 7.4

1st vibration

2nd vibration

1st vibration = 2nd vibration = 2()3rd vibration = 3()


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Newtonian Physics Breakdown-Quantization of Energy-

Max Planck

(1858-1947)Solved the

ultraviolet

catastrophe

It was believed that like wave theory,

energy was also continuous.


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Figure 7.5

Objects can gain or

lose energy by

absorbing or

emitting radiant

energy in QUANTA.

Intensity should

Increase withDecreasing . As youadd more energy,

atoms should vibrate

with a higher energy,

in a continuous

fashion.


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Energy of a vibrating system (electro-magnetic radiation) is proportional to

frequency.

Ep = h h = Plancks constant = 6.6262 x 10-34Js

Quantization of Energy

We now MUST abandon the idea that

Energy acts as a continuous wave!


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From Planck on to Einstein


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Photoelectric Effect

A. Einstein (1879-1955) Experiment demonstrates the particle

nature of light. (Figure 7.6)

Classical theory said that E of ejectedelectron should increase with increasein light intensitynot observed!

No e- observed until light of a certain

minimum E is used & Number of e- ejected depends on light

intensity.


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Experimental observations says that light

consists of particles called PHOTONShaving discrete energy.

It takes a high energy particle to bump

into an atom to knock its electron out,

hence the use of a mv

2

term. It would take some minimum energy i.e.

critical energy to knock that electron

away from its atom.


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Energy of Radiation

PROBLEM: Calculate the energy of1.00 mole of photons of red light.

= 700. nm ( c = ) = 4.29 x 1014 sec-1Ep = h

= (6.63 x 10-34Js)(4.29 x 1014 sec-1)

= 2.85 x 10-19 J/photon

Notice Einstein's use of Planck's formula.


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Energy of Radiation

Energy of 1.00 mol of photons of red light.Ep = h

= (6.63 x 10-34Js)(4.29 x 1014 sec-1)

= 2.85 x 10-19 J per photon

E per mol =

(2.85 x 10-19 J/ph)(6.02 x 1023 ph/mol)

= 171.6 kJ/mol

This is in the range of energies that can break

bonds.


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A minimum frequency is required to cause any current

flow. Above that frequency, the current is related tothe intensity of the light used. The ejected electrons(since we are talking about collisions between photonsand electrons) also have more kinetic energy when

higher frequencies are used.

EK = 1/2 meve2 = Einput - Eminimum

Einstein finds:

Ep = h = 1/2 meve2, evidence that photons have bothwave/particle properties



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Light is used to eject an electronfrom a metal. Calculate the velocity

of the ejected electron if the photon

used to eject the electron has awavelength of 2.35 x 10 -7 m and the

minimum frequency required to

eject an electron is 8.45 x 10 14 s-1.


Step by step!!


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The Final Crack in Classical,Newtonian Physics

MONUMENTAL Edifice Planck---Energy is NOT Continuous like

waves

Einstein---Energy comes in packets or isQuantized and energy also has some

wave and particle behavior

Bohr---Applies Quantized idea to atomicparticles.the H1Atom to explain..


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Atomic Line Spectra andNiels Bohr

Bohrs greatest contribution toscience was BUILDING a

SIMPLE MODEL of the ATOM.

It was based on an

understanding of the LINESPECTRA of excited atomsand its relationship toquantized energy.

Niels Bohr

(1885-1962)


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Line Spectraof Excited Atoms

Excited atoms emit light of only certainwavelengths (Planck).

The wavelengths of emitted light depend on the

element.


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Figure 7.7


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Figure 7.8


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Figure 7.9


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Visible lines in H atom spectrum arecalled the BALMER series.

High E

Short High

Low E

Long Low

Line Spectra

of Excited Atoms


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Shells or Levels!!

Why??


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Figure7.12


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Excited Atoms Emit Light


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Atomic Spectra and Bohr

1. Any orbit (like a wave-see slide 3) should bepossible and so should any energy.

2. But a charged particle would always be accelerating

from the nucleus (vector velocity is alwayschanging) and since it is moving in an electric fieldwould continuously emit energy.

End result should be destruction since the energymentioned in the previous step is finite!

+

Electron

Orbit

One view of atomic structure in early 20thcentury was that an electron (e-) traveled aboutthe nucleus in an orbit.


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Bohr said classical (Newtonian) view is wrong!!!.

Need a new theory now called QUANTUM orWAVE MECHANICS.

e- can only exist in certain discrete orbits called stationary states.

e- is restricted to QUANTIZED energy states.

Energy of state, En = - C/n2where n = quantum no. = 1, 2, 3, 4, ....

this describes the potential energy of an electron


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Only orbits where n = integralnumbers are permitted.

Radius of allowed orbitals, Rn,Rn= n

2 R0 with Ro = 0.0529 nm

Note the same equations comefrom modern wave mechanics

approach. Results can be used to explain

atomic spectra.

Energy of quantized state, En = - C/n2


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If e-s are in quantized energystates, then DE of states canhave only certain values. This

explain sharp line spectra.

n = 1

n = 2E = -C ( 1/2

2)

E = -C ( 1/12)


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Atomic

Spectraand Bohr

Calculate DE for e-

falling from high energylevel (n = 2) to low energy level (n = 1).

DE = Efinal - Einitial = - C [ (1/1)2 - (1/2)2 ]DE = - (3/4) CNote that the process is exothermic!

n=1

n=2E= -C(1/22)

E= -C(1/12)

E

N

E

RG

Y


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AtomicSpectra

and Bohr

DE = - (3/4)CC has been found from experiment and isproportional to RH, the Rydberg constant.

RHhc = C = 1312 kJ/mole. of emitted light = (3/4)C = 2.47 x 1015 sec-1and = c/ = 121.6 nmThis is exactly in agreement with experiment!

n=1

n=2E= -C(1/22)

E= -C(1/12)

E

N

E

RG

Y


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DE = Efinal - Einitial = - RHhc [ (1/nfinal2) - (1/ninitial2)]

A photon of light with frequency 8.02 x 1013 s-1

is emitted from a hydrogen atom when it de-

excites from the n = 8 level to the n = ? level.

Calculate the final quantum number state of

the electron.

Line Spectraof Excited Atoms


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Atomic Line Spectra andNiels Bohr

Bohrs theory was a greataccomplishment.

Recd Nobel Prize, 1922

Problems with theory theory only successful for H and

only 1e- systems He+, Li2+.

introduced quantum idea

artificially. However, Bohrs model does not

explain many e-systems.So, wego on to QUANTUM orWAVEMECHANICS

Niels Bohr

(1885-1962)


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Quantum or Wave Mechanics

de Broglie (1924) proposed

that all moving objects

have wave properties.

For light: E = mc2

E = h = hc / Therefore, mc = h / and for particles

(mass)(velocity) = h / ,the wave-nature of matter.

L. de Broglie

(1892-1987)

= hmv


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Quantum or Wave

MechanicsBaseball (115 g) at 100 mph

= 1.3 x 10-32 cme- with velocity =1.9 x 108 cm/sec

= 0.388 nmExperimental proofof wave properties

of electrons


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Schrdinger appliedidea of e- behaving asa wave to the problemof electrons in atoms.

He developed the

WAVE EQUATION.E. Schrdinger1887-1961



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Solution of the waveequation give a set ofmathematicalexpressions called

WAVE FUNCTIONS, .Each describes anallowed energy state ofan e-.

Quantization isintroduced naturally.

E. Schrodinger1887-1961



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WAVE FUNCTIONS, is a function of distance and two angles.Each corresponds to an ORBITAL the region of

space within which an electron is found.

does NOT describe the exact location of theelectron.

2 is proportional to the probability of finding an e- ata given point.

U t i t P i i l


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Uncertainty Principle

Problem of defining natureof electrons in atomssolved by W. Heisenberg.

Cannot simultaneously

define the position andmomentum (= mv) of anelectron.

We define e-energy exactly

but accept limitation thatwe do not know exactposition.

W. Heisenberg

1901-1976

QUANTUM NUMBERS


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QUANTUM NUMBERS

Each orbital is a function of 3 quantumnumbers:

n, l, and ml

Electrons are arranged in shells(levels)and subshells(sublevels).

n --> shell

l --> subshell

ml

--> designates an orbital within a subshell


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Symbol Values Description

n (major) 1, 2, 3, .. Orbital sizeand energy

where E = - RH

hc(1/n2)

l(angular) 0, 1, 2, .. n-1 Orbital shape

or type

(subshell)

ml(magnetic) - l..0..+ l Orbital

orientation

# of orbitals in subshell = 2 l+ 1

QUANTUM NUMBERS


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All 4 Quantum Numbers

Principle quantum number (n)

Azimuthal quantum number (l)

Magnetic quantum number (m)

Spin quantum number (s)

Atomic Orbitals the result of


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Atomic Orbitals the result ofQuantum Mechanics Calculations

Shells and Subshells


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Shells and Subshells

When n = 1, thenl

= 0 and ml= 0 .Therefore, if n = 1, there is 1 type of

subshell and that subshell has a

single orbital.(m

lhas a single value ---> 1 orbital)

This subshell is labeled s

Each shell has 1 orbital labeled s, and it

is SPHERICALin shape.

1 O bit l


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1s Orbital

2s Orbital


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2s Orbital

3s Orbital


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3s Orbital


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Atomic Orbitals the result of


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Atomic Orbitals the result ofQuantum Mechanics Calculations

p Orbitals


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p OrbitalsWhen n = 2, then l= 0 and 1

Therefore, in the n = 2 shellthere are 2 types oforbitals 2 subshells

Forl= 0 ml= 0

this is anssubshellForl= 1 m

l= -1, 0, +1

this is a p subshellwith 3 orbitals

planar node

Typical p orbital

See Figure 7.16

When l= 1, there is

a

PLANAR NODE

thruthe nucleus.


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O bit l


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p Orbitals

A p orbital

The three porbitals lie 90o

apart in space

2p Orbital


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2px Orbital

2p Orbital


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2py Orbital

2 O bit l


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2pz Orbital

3p Orbital


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3px Orbital

3p Orbital


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3py Orbital

3p Orbital


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3pz Orbital

d Orbitals


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d OrbitalsWhen n = 3, what are the values ofl?

l= 0, 1, 2and so there are 3 subshells in the shell.

Forl= 0, ml= 0

---> s subshell with a single orbitalForl= 1, m

l= -1, 0, +1

---> p subshell with 3 orbitals

Forl= 2, ml= -2, -1, 0, +1, +2

---> d subshell with 5 orbitals


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d Orbitals

s orbitals have no planarnode (l= 0) and so arespherical.

p orbitals have l= 1, andhave 1 planar node,and so are dumbbellshaped.

This means d orbitals,( l= 2) have2 planar nodes

typical d orbital

planar node

planar node

See Figure 7.16

3d Orbital


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3dxy Orbital

3d O bit l


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3dxz Orbital

3d Orbital


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3dyz Orbital

3d 2 Orbital


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3dz Orbital

3d 2 2 Orbital


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3dx - y Orbital

f Orbitals


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When n = 4, l= 0, 1, 2, 3 so there are 4subshells in the shell.

Forl= 0, ml= 0

---> s subshell with single orbital

Forl= 1, ml= -1, 0, +1

---> p subshell with 3 orbitals

Forl= 2, ml

= -2, -1, 0, +1, +2

---> d subshell with 5 orbitalsForl= 3, m

l= -3, -2, -1, 0, +1, +2, +3

---> fsubshell with 7 orbitals

Orbitals and Quantum Numbers


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n l ml

1 0 0 1s

2 0 0 2s

2 1 1

2 1 02 1 -1

2p



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n l ml

3 0 0 3s

3 1 1

3 1 0

3 1 -1

3p

3d

3 2 2

3 2 1

3 2 03 2 -1

3 2 -2

Sample Problems


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Sample Problems

1. Is it possible to have a d orbital in level 1?

2. Is it possible to have a 6s subshell?

3. How many orbitals are in a 7s sublevel?

4. How many orbitals are possible if n = 3?

5. What type of orbital has the quantumnumbers

a) n = 5, l= 2, ml= 1

b) n = 3, l= 2, ml=-1c) n = 6, l= 3, m

l= -3

No

Yes

One

9

5d

3d6f

Practice Problems


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1. Calculate the wavelength of a photon

having an energy of 2.58 x 10-18

J.2. In the hydrogen atom, which transition,

3 --> 2 or 2 --> 1, has the longerwavelength?

3. Calculate the wavelength of anobject (mass = 545 lbs) with a speed of45 miles/hour.

4. Give all possible sets of quantum

numbers for 4p, 3d, and 5s.

5. How many orbitals are in the

a. the third level? b. l= 3 sublevel?

Practice Problems Answers


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1. 7.71 x 10-8 2. 3 --> 2

3. 1.3 x 10-37 m 5. a) 9 b) 7

4. 4p nl

ml4 1 1

4 1 0

4 1 -1

Problem 4 continued on next slide.



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3d n l ml

3 2 2

3 2 1

3 2 0

3 2 -1

3 2 -2

5s 5 0 0

Sample Problem


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Sample Problem1. Calculate the frequency of light having a

wavelength of 1 x 10-7

m.

= c

1 x 10-7m . = 3.00 x 108 m/s = 3 x 1015/s

Sample Problem


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Sample Problem2. Calculate the wavelength of light having a

frequency of 1.5 x 108

hz.

= c.1.5 x 108 /s = 3.00 x 108 m/s

= 2.0 m

Sample Problem


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Sample Problem3. Calculate the frequency of light having a

wavelength of 1 x 103

nm.

= c

1 x 10-6m . = 3.00 x 108 m/s = 3 x 1014/s

Practice Problem


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Practice Problem1. Calculate the energy of a photon having a

frequency of 3 x 1015

/s.

Ep

= hEp = 6.63 x 10

-34 Js 3 x 1015/s

= 2 x 10-18 J

Practice Problem


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Practice Problem2. Calculate the frequency of light having an

energy of 2.0 x 105

J/mole.

Ep = h

3.3 x 10-19

J = 6.63 x 10-34

Js

= 5.0 x 1014 /s

2.0 X 105

J molemole 6.02 x 1023 photon

Practice Problem


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Practice Problem3. Calculate the energy of a photon with a

wavelength of 575 nm. = c

5.75 x 10-7m = 3.00 x 108 m/s = 5.22 x 1014/s

Ep = hEp = 6.63 x 10-34 Js 5.22 x 1014/s

= 3.46 x 10-19 J

Calculate the energy of the photon:


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gy p

Photon wavelength = 2.35 x 10 -7 m

= c

2.35 x 10-7m = 3.00 x 108 m/s = 1.28 x 1015/s

Ep = hEp = 6.63 x 10-34 Js 1.28 x 1015/s

= 8.49 x 10-19 J

Calculate the min. energy to eject an electron:


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gy j

Min. = 8.45 x 10 14 s-1.

Ep = hEp = 6.63 x 10

-34 Js 8.45 x 1014/s

= 5.60 x 10-19 J

Calculate the extra energy of the electron:


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gy

8.49 x 10-19 J - 5.60 x 10-19 J = 2.89 x 10-19 J

Calculate the velocity of the electron:

E = 1/2 m v2

2.89 x 10-19 J = (1/2) 9.11 x 10-31 kg v2

= 7.96 x 105 m/s

Calculate the energy of the photon:


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C cu e e e e gy o e p o o

Ep = h

= 3.20 x 104 J = 32.0 kJ

5.32 X 10-20

J 6.02 x 1023

photonphoton mole

Ep = 6.63 x 10-34 Js 8.02 x 1013/s

Calculate the level number :


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Ep = hDE = -C [(1/n)2 - (1/n)2]-32. kJ = -1312 kJ [(1/n)2 - (1/8)2]

n = 5

8-ch7(struktur atom).ppt

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