Vacuum tube

-

V

, only for shorter than certain wavelength

Current

VVo

Fixed wavelengthVarying intensity

I

2I

3I

Maximum electron energy

0

Current

V

1

2

3

1 <2 <3

Q. Why the maximum electron energydepend on the wavelength (frequency)?Why strong beams of long wavelength cannot knock out electrons?

Classical EM theory (wave theory) tells us that the max electron energy should depend on only the intensity of the light.

In 1900, Max Planck postulated that electromagnetic energy is emitted in discrete packets, or quanta. The energy of these quanta is proportional to the frequency of the radiation.

E = hf = hc/

Huge water wave, large amplitude

sound wave

h = 6.6 x 10-34 Js (Planck’s constant)

In explaining the photoelectric effect, Einstein picked up the idea of Planck, and proposed in 1905 that light was not only emitted by bundles of energy E = hf, but it was also absorbed in such bundles.

Light Quanta = Photons

Kinetic Energy of an electron = Ephoton - Ethreshold

Ephoton = hc/ > Ethreshold

Work function: depends on material ~ eV

02

1 2 mv

-

light quantum: photon

In this case, light can be considered as a massless particle, photon with energy solely determined by its wavelength (frequency).

There is no conflict with Einstein’s relativistic mechanics: massless particle can have speed of light!

E = hf = hc/h = 6.6 x 10-34 Js (Planck’s constant)

= 580 nm photon (yellow light) carries

E = (6.6 x 10-34 Js)(3 x 108 m/s)/(580 x 10-9 m) = 3.4 x 10-19 J

How do we understand the intensity of light?

Intensity of light (EM radiation) ≈number of photonswith same wavelength or frequency

Wave behaves like a particle.Does a particle behaves like wave?

It would seem that the basic idea of the quantum theory is the impossibility of imagining an isolated quantity of energy without associating with it a certain frequency

de Broglie in 1923 as a graduate student

light quantum (photon) E = hf = hc/

Relativistic case

2242 cpcmE o

massless

E = pc = hc/ = h/pwave quantity

particle quantity

for mo = 0

= h/p

De Broglie proposed that all particles (electrons) should have wavelength associated with their momentum in exactly the same manner.

particle

wave

Q. What is the de Broglie wavelength of an electron that has a kinetic energy of 100 eV?

After an electron is accelerated in 100 V potential difference, its kinetic energy is 100 eV.eV unit has to be converted into SI unit, Joule.

1 eV = 1.6 x 10-19 J

Ek = (1/2)mov2 = 1.6 x 10-17 J

v2 = 2Ek/mo = 2(1.6 x 10-17 J)/(9.1 x 10-31 kg) = 3.52 x 1013 m2/s2

v = 5.93 x 106 m/s low speed: no need to use relativistic

= h/p = h/mov = (6.6 x 10-34 Js)/(9.1 x 10-31 kg x 5.93 x 106 m/s) = 1.23 x 10-10 m = 0.123 nm

C.J. Davisson and L.H Germer scattered low energy electrons off The Ni crystal and observed a peculiar pattern of scattered electrons.

Diffraction pattern from the regular Crystalline structure of order of 0.1 nm size.

Hydrogen Spectrum

Emission SPECTRUM

ABSORPTION SPECTRUM

Bohr’s Hydrogen Atom and the Electron as a Wave

Niels Bohr (1885 – 1962) Louis de Broglie (1892 – 1987)

In 1913, Rutherford’s atom received a quantitative description from Niels Bohr who explained experimentally observed discrete nature of atomic spectrum of Hydrogen. In spite of its immediate success in providing theoretical account of the spectrum and other nature of Hydrogen atom, a complete understanding of Bohr’s atom came only after de Broglie’s conjecture (1923) that electrons should display wave properties.

+

-

Bohr’s Hydrogen Atom(1 proton and 1 electron)

+

- - -

2

22

r

ek

r

mv

centripetal force = Coulomb force

electron as a wave(de Broglie)

2r = , 2, 3, … = h/mv

This is the necessary condition for electron to maintain an orbit.

2

22

r

ek

r

mv

2r = , 2, 3, … = h/mv

2rn = nh/mvn n

mke

hnrn 22

22

4

nh

kevn

22

rn = (5.3 x 10-11)n2 (m); vn = 2 x106/n (m/s)

nn rrn n (nm)(nm) vvnn (m/s) (m/s)11 0.0530.053 2 x 102 x 1066

22 0.2120.212 1 x 101 x 1066

33 0.4770.477 5 x 105 x 1055

2 x 0.053 0.1 nm is the size of Hydrogen atom

Now, let’s think about the total energy of the electron in the nth orbit.

Etot = Potential Energy + Kinetic Energy-ke2/rn(1/2)mvn

2

mke

hnrn 22

22

4

nh

kevn

22

eVnhn

mekE ntot 222

422 6.132

nn EEn n (eV)(eV)

11 -13.6-13.6

22 -3.4-3.4

33 -1.5-1.5

∞∞ 00

nh

kevn

22

mke

hnrn 22

22

4

+

- - -

Etot

0 n = ∞

-13.6 eV n = 1

n = 2-3.4 eV

n = 3-1.5 eV

-

eVnhn

mekE ntot 222

222 6.132

Ionized state of Hydrogen: proton

Electron Energy diagram of Hydrogen atom

Etot

0 n = ∞

n = 2-3.4 eV

n = 3-1.5 eV

eVnhn

mekE ntot 222

222 6.132

n = 1-13.6 eV -lowest energy:ground state

Electron has to absorb 10.2 eV energy for this

-

Electron has to absorb 12.1 eV energy for this

eVnhn

mekE ntot 222

222 6.132

22

116.13)(

nmEEmnE mn in eV

17

22

10097.1

111

mR

nmR

mn (Rydberg const.)

E(nm) or < 0 Absorb photons of a given E (nm) or > 0 Emit photons of a given

E = hc/

Etot

0 n = ∞

n = 2-3.4 eV

n = 3-1.5 eV

n = 1-13.6 eV -

E (12) = 10.2 eV = 10.2 x (1.6 x 10-19 J/eV) = 1.63 x 10-18 J

E = hc/

= 121 nm

Ultraviolet range

Q What is the longest wavelength em radiation that can ionizeunexcited hydrogen atom?

smallest energyground state (n = 1)

n = 1 m = ∞

17

22

10097.1

111

mR

nmR

mn

1/ = (1.097 x 107) x (0 – 1) = - 1.097 x 107 (m-1)

= -9.12 x 10-8 (m) = -91.2 (nm) (- means absorption)

2

1

11

1

mR

m

22

2

1

2

11

mR

m

22

3

1

3

11

mR

m

Q. What is the shortest wavelength for the Balmer series?

Largest energy difference in Balmer seriesFrom n = ∞ to m = 2 (Balmer)

17

22

10097.1

111

mR

nmR

mn

= (R x (1/4 – 0))-1

= 365 nm

The Balmer series are in the visible range.