3 - Electronic structure (energy) class version.ppt

Electronic structure

Electronic structure of atoms

Where are the e

s located and

How can we probe an atom to find

out?

Light

Electromagnetic radiation

Is the energy in the form of light which travels

through space as waves at a velocity of

~3.0 x 10

m/s (the speed of light)

A small part of the Electromagnet

Spectrum

The dr Broglie Hypothesis

Louis de Broglie said that all matter

has wave characteristics

This is important because sometimes

the behavior of electrons is better

described in terms of waves instead of

particles

light

Considered an Electromagnetic Wave

Waves can build each other up or cancel

each other – Interference

Waves can bend around edges –

Diffraction

Waves

All forms of electromagnetic radiation exhibits wave properties

Velocity, wavelength, amplitude, and frequency

Velocity (c) = 3.00 x 108 m/s in vacuum (speed of

light)

Wavelength (λ) = lambda - distance from start of one wave to start of next wave (nm)

Shorter the wavelength, the stronger the energy

Period (T) = Time for one wave to go by (seconds)

Frequency (f or v) = nu - Number of waves

passing by in one second (hz) = 1/T

Wavelengths and Visible light

Red low energy, longest wavelength

Orange Yellow Green Blue Indigo

waves

Speed of wave (c) = ?

c = λ f

C = speed of light

(c) = 3.00 x 10

m/s in vacuum

If λ = 550 nm then f = ?

If f = 7.52 x 10

Hz then λ = ?

Conversion we need

c = λ f

C = speed of light

(c) = 3.00 x 10

m/s in vacuum

If λ = 550 nm then f = ?

c = λ f

3.00 x 108 m 1 x 109 nm

--- x --- = 3.00 x 1017 nm

1 1 m

3.00 x 1017 nm/s = 550 nm (f)

c = λ f

C = speed of light

(c) = 3.00 x 10

m/s in vacuum

If f = 7.52 x 10

Hz then λ = ?

c = λ f

3.00 x 10

m/s = λ (7.52 x 10

1/sec)

= 3.99 x 10

m

3.99 x 10

m 1 x 10

nm

--- x --- = 3.99 x 102 nm

1 1 m

Light

Considered as “particles”

Photons or Quantas

Intensity or Power of light depends on the

height of the wave or number of photons

Energy of light though depends on the

frequency/wavelength

Red light (750 nm) does not expose film

Violet light (450 nm) does expose film and

causes sunburns

Electrons and energy

The positively charged nucleus is

always pulling at the negatively

charged electrons around it

This gives the electrons potential energy

that increases with their distance from

the nucleus

like gravitational potential energy of a brick on the third floor of a building

This is greater than if the brick was on the first floor

Energy of an electron is

quantized

Remember from electron configuration –

energy levels are fixed

Electrons can only exist at specific energy

levels, separated by specific intervals – the

rows

Electrons exist at ground (normal)

configuration most of the time

When heat is absorbed, they become

excited

The electrons have to emit the same amount

of energy absorbed to return to ground state

Quantized energy of an electron

in a hydrogen atom

Can be found if you know its principal

quantum number or shell (row

number)

Energy of an electron

E

= -2.178 x 10

/ n

joules

En = the energy of the electron

n = the principal quantum number of the electron

Electromagnetic radiation

When atoms absorb energy in the form of

electromagnetic radiation, electrons jump to higher energy levels

When electrons drop from higher to lower energy levels, atoms give off energy in the form of

electromagnetic radiation (light)

The energy level changes for the electrons of a

particular atom are always the same, so atoms can be identified by their emission and absorption spectra The relationship between the change in energy level of an electron and the electromagnetic radiation

absorbed or emitted is given by the equation:

Energy of light

E = hf or (hv)

h = Planck’s constant = 6.63 x 10-34 J/Hz

V = frequency

E = energy of a quantum of light (photon)

Since c = λ v

Then we can say that v = c / λ Therefore, E = hc/λ

H = Plank’s constant C = speed of light

Energy and electromagnetic

radiation



E = hv = hc /

λ



h = Planck’s constant, 6.63 x 10-34 joule * sec

v = frequency of radiation λ = wavelength of radiation

Practice

What is the frequency of red light that

has a wavelength of 700.0 nm? What

is the energy of a photon of this light?

What is the frequency of violet light

that has a wavelength of 410.0 nm?

What is the energy of a photon of this

light?



What is the frequency of red light that

has a wavelength of 700.0 nm? What

is the energy of a photon of this light?



E = hv = hc / λ

v = c / λ

v = 3.00 x 10

m/s / 700.0 nm

v = 3.00 x 10

nm/s / 700.0 nm

v = 4.29 x 10

Hz

E = hv

E = 6.63 x 10

joule * sec (700.0 nm)

There is a simple relationship

between an electrons wave and

particle characteristics

The de Broglie Equation

λ = h/mv

λ = wavelength associated with a particle m = mass of the particle

v = speed of the particle

mv = p = momentum of a particle

H = Planck’s constant = 6.63 x 10-34 J/Hz

This hypothesis is useful for very small particles, such as electrons. For larger particles, the wavelength becomes too small to be of interest

Atomic Structure

Photon given off when

electron moves from higher

level to lower level.

Energy emitted

Hydrogen Atom

Ryberg discovered that the energy of each

emission is related to a whole number multiple

“n” (row number)



E = -R (1/n

- 1/n

) , R = 2.180 x 10

J

n=1 n=2 n=3 n=4 n=5

Another way of looking at this

hv

= -R

(1/n

- 1/n

)

Or

v = R

/H (1/n

- 1/n

)

R = 2.180 x 10

J

example

Calculate the wavelength in nanometers of the

line in the Balmer series that results from the

transition n= 3 to n = 2.

v = R

/H (1/n

- 1/n

)

v = (2.180 x 10

J) ( 1/(2)

– 1/(3)

)

6.626 x 10

Js

v = 4.570 x 10

1/s

Since λ

= c/v = 3.0 x 10

m/s /

4.570 x 10

1/s

You try one

Calculate the wavelength in nanometers of the

line in the Balmer series that results from the

transition n= 2 to n = 1.

v = R

/H (1/n

- 1/n

)

v = (2.180 x 10

J) ( 1/(1)

– 1/(2)

)

6.626 x 10

Js

v = 2.47 x 10

1/s

Since λ

= c/v = 3.0 x 10

m/s /

2.47 x 10

1/s

Atomic spectra

Blackbody radiation

• Continuous radiation distribution

• Depends on temperature of radiating object

• Characteristic of solids, liquids and dense gases

Line spectrum

• Emission at characteristic frequencies

• Diffuse matter: incandescent gases

• Illustration: Balmer series of hydrogen lines

Bell work

Calculate the energy that can be

emitted or absorbed at a wavelength

of 645 nm.

What frequency and wavelength of

radiation has photons of energy 8.23

x 10

J? What type/color of

electromagnetic radiation is this (refer

to EM spectrum)?

Answer to homework

End of chapter 6 work #2

A) c = λ f

If λ = 423 nm then f = ?

– c = λ f

3.00 x 108 m 1 x 109 nm

--- x --- = 3.00 x 1017 nm

1 1 m

= 3.00 x 1017 nm/s = 423 nm (f)

F = 7.09 x 1014 hz

B) E = hv

E = 6.626 x 10-34 J*s (7.09 x 1014 s-1)

E = 4.70 x 10-19 J

C) 1 mole of photons = 6.02 x 10 23 photons

4.70 x 10-19 J 1 kJ 6.02 x 10 23 photons

--- x --- x --- = 283 kJ/mol 1 photon 1000 J 1 mol of photons

#4

A) c = λ f

If λ = 1498 nm then f = ?

– c = λ f

3.00 x 108 m 1 x 109 nm

--- x --- = 3.00 x 1017 nm

1 1 m

= 3.00 x 1017 nm/s = 1498 nm (f)

F = 2.00 x 1014 hz

Infrared

B) see above C) E = hv

E = 6.626 x 10-34 J*s (2.00 x 1014 s-1)

#6

c = λ f

If λ = 643 nm then f = ?

– c = λ f

3.00 x 108 m 1 x 109 nm

--- x --- = 3.00 x 1017 nm 1 1 m

= 3.00 x 1017 nm/s = 643 nm (f)

F = 4.67 x 1014 hz

B) E = hv

E = 6.626 x 10-34 J*s (4.67 x 1014 s-1)

Homework

10.

V = R

/h (1/(n

)

– 1/(n

)

)

V = 2.180x10

J / 6.626x10

J*s * (1/

(2)

– 1/(4)

)

V = 3.290x10

Hz (1/4 – 1/16)

12.