An Introduction to Quantum

SPH4U

Recall that “blackbodies” (objects that emit and absorb the full spectrum) have an intensity peak in their emission spectrum corresponding to their temperature.

However, nothing in classical physics explained why objects could not emit lots of very small wavelength radiation.

This was known as the ultraviolet catastrophe.

Max Planck hypothesized that when light was emitted, it was not emitted continuously but in packets called quanta

Max Planck hypothesized that when light was emitted, it was not emitted continuously but in packets called quanta and that the energy of a single quantum is inversely proportional to the wavelength (or directly proportional to the frequency).

fEorE 1

For a single quantum:

hfE

h is Planck’s constant

sJh 341063.6

For a single quantum:

The total energy must be an integral multiple:

hfE

h is Planck’s constant

nhfE

sJh 341063.6

Electron Volts

Since the energies are very small, they are often measured in electron volts (eV).

Electron Volts

Since the energies are very small, they are often measured in electron volts (eV).

Consider the change in energy of an electron moved through a potential difference of 1 V:

eVJVCqVW 110602.1110602.1 1919

Example

Calculate the energy, in Joules and electron volts, of red light with a wavelength of 633 nm.

Example

Calculate the energy, in Joules and electron volts of red light with a wavelength of 633 nm.

?

1063.6

100.3

10633

34

8

9

E

sJh

c

m

sm

Example

Calculate the energy, in Joules and electron volts, of red light with a wavelength of 633 nm.

Example

Calculate the energy, in Joules and electron volts, of red light with a wavelength of 633 nm.

JorJE

m

sJE

hcEfcandhfE

sm

1919

9

834

1014.3101422.3

10633

1000.31063.6

Example

Calculate the energy, in Joules and electron volts, of red light with a wavelength of 633 nm.

eVJ

eVJE 96.1

10602.1

1101422.3

1919

Earlier, Hertz had discovered that certain negatively-charged metallic surfaces lost their charge (ejected electrons) when exposed to high-frequency light.

This was called the photoelectric effect.

The maximum energy of the ejected electrons could be determined by applying a retarding potential and measuring when the current dropped to zero (the cutoff potential).

If the frequency was below some threshold frequency, no electrons would be emitted.

Above this threshold, the resulting current was proportional to the intensity of the light.

The threshold frequency was different for different materials.

The higher the frequency of the light, the greater the energy of the emitted electrons.

Classical physics could not explain this or why lower frequency light did not result in any energetic electrons even at high intensities (measured in Watts).

It was Einstein who uses Planck’s light quanta (which he called photons) to explain the photoelectric effect.

Electrons near the surface of the metal can absorb the energy of incident photons.

It was Einstein who uses Planck’s light quanta (which he called photons) to explain the photoelectric effect.

Higher energy (higher frequency) photons give the electron enough of a kick to free it from the metal.Excess energy is carried away as kinetic energy.

It was Einstein who uses Planck’s light quanta (which he called photons) to explain the photoelectric effect.

Even if there are a lot of lower energy photons (high intensity light), the individual kicks are too small.

Einstein’s Photoelectric Equation

WhfEk kinetic energy of emitted electron

work function (minimum energy to eject an electron) for the material

Example

Light with a wavelength of 6.00 x 10-7 m is directed at a surface with a work function of 1.60 eV. Calculate:

(a)the maximum kinetic energy, in Joules, of the emitted electrons.

(b) their maximum speed.(c) the retarding potential needed to stop these

electrons.

Calculate (a) the maximum kinetic energy, in Joules, of the emitted electrons.

?

1056.21

10602.160.1

1063.6

1000.3

1000.6

1919

34

8

7

k

sm

E

JeV

JeVW

sJh

c

m

Calculate (a) the maximum kinetic energy, in Joules, of the emitted electrons.

JE

Jm

sJE

Whc

WhfE

k

sm

k

k

20

197

834

1055.7

1056.21000.6

1000.31063.6

Calculate (b) their maximum speed.

?

1011.9

1055.731

20

v

kgm

JE

e

k

Calculate (b) their maximum speed.

?

1011.9

1055.731

20

v

kgm

JE

e

k

sm

kk

v

kg

Jv

m

EvmvE

5

31

20

221

1007.4

1011.9

1055.72

2

Calculate (c) the retarding potential needed to stop these electrons.

?

10602.1

1055.7

0

19

20

V

Cq

JE

e

k

Calculate (c) the retarding potential needed to stop these electrons.

?

10602.1

1055.7

0

19

20

V

Cq

JE

e

k

VVC

JV

q

EVqVE k

k

472.010602.1

1055.7

0

19

20

0

00

More Practice

Textbook Questions• p. 597 #2• p. 598 #3, 4• p. 604 #12 – 15