## Definition

The previous article explained, in general terms, dealt with what quantum physics is, some applications, and examples. Herein we will discuss the **photoelectric effect** and applications governing the **double slit experiment**, the **uncertainty principle**, and the **quantum particle**.

Photons are the building blocks of light. Whether radio waves or ultraviolet waves, they are essentially photons. Unlike light, a collection of photons, individual photons have three properties: **particle**, **wave**, and **excitation**.

Photons do not have mass and, thus, can travel at the **speed of light** in a vacuum at 3 x 10^{8} m/s.

## How classical physics fails

In describing photons fully, classical physics – as we define it as – fails. I will discuss some areas where classical physics initially professed to explain behaviour but failed experimentally:

In the **photoelectric effect**, photons exhibit alternate behaviour forcing us to adopt their nature as particles. In the **Compton effect**, X-ray behaviour cannot be reliably identified by the means known to classical physics.

We will explore these two cornerstone phenomena next:

## The photoelectric effect

An electron is emitted when a beam of light shines onto certain metals. This phenomenon is known as the *photoelectric effect, *and the emitted electrons are referred to as *photoelectrons*.

To observe the photoelectric effect in practice, we need:

- a vacuum chamber
- a metallic object
- a detector
- a battery
- an ammeter

In the above experiment, a vacuum chamber is connected to the battery. Within the vacuum chamber are the detector (connected to the *positive *end of the battery) and the metal surface (connected to the **negative** end of the battery). For the plate to release an electron, the incident light must have a higher **threshold frequency** than the metallic plate’s.

**Classical prediction**

In classical physics, when we increase a wave’s amplitude, we expect its energy to increase and, in effect, its brightness. When we have blue light and increase the intensity at which the blue light falls onto a surface, we expect the surface electrons to vibrate more readily. But does this happen?

**Experiment results**

Instead, what the researchers found with the experiment was pertinent in the following diagrams.

**High Intensity/High Frequency**

**Low Intensity/Low frequency**

**High Intensity/Low frequency**

**Low Intensity/High frequency**

The following factors seem to directly correlate to how frequent electron emission takes place in a given time:

● An increased intensity increased the electron count (current), but not its speed (kinetic energy)

● An increased frequency increased the speed of the electrons, but not the amount. (Review the graphs above)

Given enough time, it is evident that all surface electrons would emit.

**More information on the photoelectric effect**

Whether an electron is released depends on the **work function.** Suppose the incident light’s velocity is such that it can overcome the surface’s work function. In that case, the electron is released and – as every object with motion – will possess kinetic energy, E_{k} Only as a particle can the photoelectric effect describe the properties of light. Einstein’s formula shows this relationship:

*E=hf *

where *h* is the Planck constant and *f* is the oscillating frequency.

From what we know in Physics 1, when an object changes in momentum, the momentum is *conserved*. Thankfully, this also applies to the quantum world. But the formula is a tad-bit different. Recall that E_{(total) }= E_{k} + E_{p }and that kinetic energy

We can surely replace the potential energy E_{(total)} with the total energy *hv*, and the potential energy with the electron’s **potential energy **(its mass and velocity) in Einstein’s formula. The end result will look like this:

where is the threshold frequency and the frequency of incident light.

From Einstein’s formula, we can deduce that:

- the energy is proportional to the frequency ;
- the independent side (right-hand side) is scaled by the Plank constant ;
- the brightness is increased by increasing the intensity of light (number of photons), and thus increases the electrons emitted;
- both the wave and particle models complement each other (the
**principle of complementarity**).

## The Compton effect

Another example of how classical physics fails to predict the behaviour of photons is the *Compton effect*. Light has a particle nature is evident when photons (i.e., in an X-ray) interact with free electrons. Because the resultant photon changes trajectory and transfers its energy to the electron, the photon will now have a different wavelength, and also different energy.

Here is an illustration of what is happening in an X-ray:

Albert Einstein showed how a photon of energy possessed a momentum of *p = E/c = hf/c*.

The **Compton shift equation** is given as follows:

where the **electron’s mass is** m_{e }and the **Compton wavelength **is the factor h/m_{e}c.

Notice how this formula plays on the old formulas from classical physics: in the spirit of momentum, m_{e}c is similar to mv^{2}, where m_{e } is the electron’s mass, and c is the speed of light.

## The wave properties of particles

Recall that from Physics 1, that frequency is inversely proportional to wavelength, otherwise written as:

or

It necessarily follows that the higher the frequency (and the shorter the wavelength), the higher the energy at which the wave fluctuates.

where mv is the product of the mass, *m*, and frequency, *v*, which can also be written as momentum *p*.

## The quantum particle

Light is said to have a dual nature – either that of a tennis ball or of an ocean wave. That is how we characterize it in classical physics. We put it into easy boxes that we can understand; However, in the study of Quantum Physics, there are different shades of grey. Contradictions might lull us into a rabbit hole: sometimes the very wave-like forms now exhibit particle-like behavior. Because of the nature of both particle and wave models, physicists now refer to this dual behavior as the **quantum particle**.

**Wave properties:**

- Infinite length-
**unlocalized** - Single frequency

**Particle properties:**

- Zero mass
- Local to a specific point in space –
**localized**

### The effects of superimposition

Imagine that we have 2 waveforms, *wave 1* and *wave 2,* and we want to analyze their behavior when occurring together. We can represent its interference below as it adds up nicely:

**Wave 1**

**Wave 2**

This adds up to the following (a wave having twice the amplitude of the 1st and/or the 2nd):

These are ideal waves. But what if 1 wave is out of phase (meaning that 1 wave is now shifted a bit along the x-axis)? Then we have something not so pretty called *beats*.

**Wave 1**

**Wave 2 **(now wave 1 with a shift)

Notice the odd shape in the result. You may imagine what we will give once we add more waves. This would be what physicists refer to as a *wave packet*.

For individual waves, we can get the speed by the formula . This is called the **phase speed **because it is the rate (how quickly) the crest of a wave moves in a given direction.

## Heisenberg’s uncertainty principle

Quantum Physics has it that to measure position and momentum simultaneously, we enter into the world of contradictions. This is known as *Heisenberg’s uncertainty principle*. In short, Heisenberg’s uncertainty principle is the product of 2 uncertainties of position and momentum, which is more significant than some value . We can formulate this principle in 2 ways: *first*, by introducing *momentum *and *position*; and *second*, by introducing *energy *and *time*:

which relates to the form with energy and time, . Given that . We can also write the above formulas as follows:

and respectively.

**Example Problems **

**Problem 1 **

An electron travels at a speed of 6 x 10^{3} m/s and has an accuracy of 0.003%. Calculate the **minimum value** for the uncertainty in determining the electron’s position.

**Solution**:

(i). First, show the momentum .

(ii). Next, calculate the uncertainty

**Problem 2 **

Calculate the line width using the uncertainty principle given that .

**Solution **

(i). First, recall the formula relating energy to frequency *E=hf *

(ii). Next, we can rearrange this to get frequency alone, looking at the change in frequency and change in energy .

(iii). Next, we can rearrange the uncertainty principle for the change in energy .

(iv). Next, we can substitute which should give us .

(v.). Finally, we can substitute the value for given into the above formula for (iv) and will get: .