Quantum Physics (work in progress)

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In 1905, Albert Einstein published a
paper that described his photon
theory of light. This explained
experimental evidence, such as the
photoelectric effect, that did not fit
with the classical wave model of light.

Einstein proposed that, although light
exhibited wave-like properties, it
travelled in particles called photons.

He explained that photons contain discrete ‘energy packets’
called quanta, and that the energy of an individual quantum
depends on the frequency of the light.

Photon energy

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Thermal radiation

All objects with a temperature above absolute zero emit
thermal radiation.

Why do objects emit this radiation?

All charged particles emit
radiation when they accelerate.

If an object has a temperature
above absolute zero, then its
electrons and protons will
vibrate and emit radiation.

Absolute zero is the temperature (-273°C, 0K) at which
particles theoretically lose all their energy and stop vibrating.

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Incandescence

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The incandescent light bulb

How could you use incandescence to explain how a
conventional light bulb works?

1. A current flows through the filament.

2. The filament heats up.

3. The thermal emission of the filament

moves into the visible spectrum.

Why is the incandescent light bulb not an efficient device
for producing light?

The filament’s emission spectrum remains mostly in the
infrared, even when it is at its hottest. Most of the bulb’s
energy input is therefore wasted as heat.

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Emission spectra

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Explaining the origin of line spectra

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Dropping energy levels

Each possible drop between atomic energy levels in an
atom corresponds to the emission of one specific frequency
of photon.

This results in one line in
an element’s emission
spectrum. Three lines in
hydrogen’s spectrum are
shown here, along with the
energy jumps that they
correspond to.

hydrogen

n = 1

n = 2
n = 3
n = 4

ground

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The fluorescent lamp

The incandescent light bulb is an inefficient device for
producing light (typically 1–2%). The fluorescent lamp is
an alternative, with an efficiency of around 10%.

Fluorescent lamps use electricity to
cause excitation of mercury
vapour. When the mercury atoms
relax, they emit ultraviolet photons.

Ultraviolet light is not visible, but can
be converted into visible light using a
phosphor. This coats the inside of
the bulb and fluoresces when
bombarded with the ultraviolet light
from the mercury vapour.

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The fundamental mechanism for the conversion of electrical energy to light
is the emission of a photon when an electron in a mercury atom falls from
an excited state into a lower energy level. Electrons flowing in the arc collide
with the mercury atoms. If the incident electron has enough kinetic energy, it
transfers energy to the atom's outer electron, causing that electron to
temporarily jump up to a higher energy level that is not stable.

The atom will emit an ultraviolet photon as the atom's electron reverts to a
lower, more stable, energy level.

These are not visible to the human eye, so ultraviolet energy is converted to
visible light by the fluorescence of the inner phosphor coating.

The difference in energy between the absorbed ultra-violet photon and the
emitted visible light photon heats the phosphor coating.

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Why mercury?

Humphrey Davy discovered the carbon
arc(h) lamp at the start of the 19^th century.

In 1821 he demonstrated the mercury
vapour arc lamp at the Royal Institution.

Boiling point of Mercury is 357 ^oC, the glass tubes operate at around 500 ^oC

Faraday’s motor used mercury,
since it was liquid at room temp.

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Why does the light appear white?

Most of the photons that are released from the
mercury atoms have wavelengths in the UV region
of the spectrum, predominantly 253.7 and 185 nm.

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Without mercury v without phosphor coating

Why does the light appear white?

A fluorescent lamp tube is filled with a
mix of argon, xenon, neon, or krypton,
and mercury vapor. The pressure
inside the lamp is around 0.3% of
atmospheric pressure. The partial
pressure of the mercury vapor alone is
about 0.8 Pa (8 millionths of
atmospheric pressure), in a T12 40-
watt lamp.

The inner surface of the lamp is
coated with a fluorescent coating
made of varying blends of metallic
and rare-earth phosphor salts.

The lamp's electrodes are typically
made of coiled tungsten and are
coated with a mixture of barium,
strontium and calcium oxides to
improve thermionic emission.

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A typical high pressure mercury lamp

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Why argon?

Electric current flows through the tube in a low-pressure arc discharge.
Electrons collide with and ionize noble gas atoms inside the bulb
surrounding the filament to form a plasma by the process of impact
ionization. As a result of avalanche ionization, the conductivity of the ionized
gas rapidly rises, allowing higher currents to flow through the lamp.

The fill gas helps determine the electrical characteristics of the lamp but
does not give off light itself. The fill gas effectively increases the distance
that electrons travel through the tube, which allows an electron a greater
chance of interacting with a mercury atom.

Additionally, argon atoms, excited to a metastable state by the impact of an
electron, can impart energy to a mercury atom and ionize it, described as
the Penning effect. This lowers the breakdown and operating voltage of the
lamp, compared to other possible fill gases such as krypton.

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Quantised absorption

Just as an electron can drop between energy levels in an
atom, releasing a single photon, it can also jump up one or
more energy levels if it absorbs a photon of the right energy.

Only a single photon of the relevant energy can cause this.
It is not possible for an electron to ‘store up’ energy from
smaller quanta until it has enough to make the jump.

One result of this is that shining
a continuous spectrum of light
at a transparent material leads
to a few discrete frequencies
being absorbed, while the rest
are transmitted. This forms an
absorption spectrum.

H emission spectrum

H absorption spectrum

Hotspot

Identify the unknown gas by its absorption spectrum.

Hotspot

Identify the elements contained in the observed spectrum.

Hotspot

Identify the unknown gas from its emission spectrum.

Hotspot

Identify the unknown gas from its emission spectrum.

Hotspot

Identify the elements in the observed spectrum.

Hotspot

Identify the elements in the observed spectrum.

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Analysing light from a source

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Energy of photons in light

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Energy of a photon calculations

1.

Find the frequency of a wave with photons of energy 3.58 x 10-19J

2. Find the energy of a photon in a wave of frequency 4.8 x 1014Hz

3. Calculate the energy of red light of frequency 450 THz

4. Calculate the energy of green light of frequency 580 THz

5. Find the frequency of a wave in which the photon energy is 3.7 x 10-19J

6. Calculate the frequency of a wave with a photon energy of 2.8 x 10-19J

Math Response

Find the frequency of a wave with photons of energy 3.58 x 10^-19 J

Type answer here

Correct Answer: 5.4\times10^{14}

Deg^°

Rad

Match

Match the following

• Find the frequency of a wave with photons of energy 3.58 x 10^-19 J

• Find the energy of a photon in a wave of frequency 4.80 x 10¹⁴ Hz

• Calculate the energy of red light of frequency 450 THz

• Calculate the energy of green light of frequency 580 THz

• Find the frequency of a wave in which the photon energy is 3.70 x 10^-19 J

5.60 x 10¹⁴ Hz

5.40 x 10¹⁴ Hz

3.85 x 10^-19 J

2.98 x 10^-19 J

3.18 x 10^-15  J

• 5.60 x 10¹⁴ Hz

• 5.40 x 10¹⁴ Hz

• 3.85 x 10^-19 J

• 2.98 x 10^-19 J

• 3.18 x 10^-15  J

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The electronvolt

A joule (J) is a large unit of energy when dealing with tiny
atoms. Scientists often use an alternative unit for small
amounts of energy, called an electronvolt (eV).

How many electronvolts to one joule?

1 eV = 1.6 × 10^-19J

1J = 1/(1.6 × 10^-19)eV = 6.25 × 10¹⁸eV

Use these two conversion rates to change between the two.
Be careful to use joules in calculations with other SI units.

An electronvolt is equal to the amount of energy transferred
to a single electron if it is accelerated through a potential
difference of 1 V:

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Wave speed, wavelength and frequency

The speed, wavelength and frequency of a wave are related
by the following equation:

wave speed = frequency × wavelength

c = fλ

What is the photon energy, in electronvolts, of red
light of wavelength 685nm?

E = hf = hc/λ = 6.63 × 10^-34× 3 × 10⁸/ (685 × 10^-9)

= 2.9 × 10^-19J

= 2.9 × 10^-19× 6.25 × 10¹⁸eV

= 1.8eV

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Wave and photon calculations

1. What is the photon energy of electromagnetic radiation with a
wavelength of 5.75 x 10^-12m?

2. What is the photon energy of electromagnetic radiation with a
wavelength of 470 nm?

3. What is the wavelength of electromagnetic radiation with a
photon energy of 2.98 x 10^-24J?

4. What is the wavelength of electromagnetic radiation with a
photon energy of 6.03 x 10^-33J?

Isaac Physics

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Isaac Physics

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Isaac Physics

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Wavelength of emitted radiation

Given a set of values for atomic energy levels in a particular
element, it is possible to calculate the wavelengths of
radiation it can emit or absorb.

hf = E1 – E2

hc/λ = E1 – E2

The difference between two
energy levels gives the
energy of the photon
corresponding to that jump,
and this can be used to find
frequency and wavelength:

-0.85eV

-1.5eV

-3.4eV

-13.6eV

1

2

3

4

5

6

What wavelengths of light are emitted in transitions 1–6?

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Photoelectricity was first
discovered in 1887 by
Heinrich Hertz during
investigations into radio
waves using a ‘spark gap’.

Radio waves are produced
when a high voltage is supplied
across two electrodes causing a
spark in the gap.

Hertz found that if ultraviolet light was shone on the
electrodes, the sparks were much stronger and thicker.

The discovery of photoelectricity

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In the late 19thcentury, scientists
used apparatus like this
photocell to analyse and
measure photoelectricity.

When light is shone on the cathode
of the photocell, electrons are
released. They are attracted to the
anode, causing a current to flow.

Measuring photoelectricity

anode

cathode

vacuum

photocell

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The gold leaf electroscope

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Einstein’s photoelectric equation

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The photoelectric effect equation

1.

Calculate the frequency of incident radiation, if an electron is emitted with
an energy of 3.0 eV from a metal surface of work function 3.5 eV

2.

Calculate the frequency of incident radiation, if an electron is emitted with
an energy of 5.0 eV from a metal surface of work function 3.2 eV

3.

Calculate the wavelength of incident radiation if the electron is emitted with
an energy of 5.0 eV from a metal surface of work function 3.5 eV

4.

Calculate the wavelength of incident radiation if the electron is emitted with
an energy of 10 eV from a metal surface of work function 3.5 eV

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Wave properties of particles

In 1924, Lois De Broglie came up with a radical new way of
looking at the relationship between waves and particles. He
suggested that all particles could behave as waves.

De Broglie deduced that a particle had a
wavelength, and it was dependent on only
one thing – the momentum of that particle:

λ = h / p

Three years later, this
hypothesis was
confirmed for
electrons with the first
observations of
electron diffraction.

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De Broglie wavelength calculations

1. Find the wavelength of an electron travelling at 4.5 x 10⁷m/s if its

mass is 9.1 x 10^-31kg.

2. Find the wavelength of a man travelling at 5 m/s if his mass is 70 kg

3. Find the mass of a particle, if it has a wavelength of 3.2 x 10^-21m

and is travelling at 2.0 x 10⁶m/s.

4. Find the velocity of a particle if it has a wavelength of 3.6 x 10^-21m

and a mass of 4.2 x 10^-10kg.

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Waves and particles

Drag and Drop

All particles exhibit wave-like properties

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true

false

Drag and Drop

All waves are also particles​

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true

false

Drag and Drop

The de Broglie wavelength of a particle can be found with the equation: λ = ​ hp​

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true

false

Drag and Drop

The photoelectric effect is a demonstration of the particulate nature of light​

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true

false

Drag and Drop

Electron diffraction is a demonstration of the particulate nature of electrons​

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false

true