All you need to know about Quantum Physics

Table of Contents


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 108 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.

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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

All the images above are by  

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, Ek Only as a particle can the photoelectric effect describe the properties of light. Einstein’s formula shows this relationship:

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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) = Ek + Ep and that kinetic energy

E_{k} = \frac{1}{2}mv^{2}

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:

K (max) = e \Delta V_{s}

hv = e V_{0} + \frac{1}{2mv^{2}} where V_{0} is the threshold frequency and v 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:

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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:

Image by hyperphysics.phy
\lambda ^{^{\acute{}}}- \lambda _{n}= \frac{h}{m_{e}c^{1-cos\theta }}

where the electron’s mass is me and the Compton wavelength is the factor h/mec.

Notice how this formula plays on the old formulas from classical physics: in the spirit of momentum, mec is similar to mv2, where me 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:

f \alpha \frac{1}{\lambda } or \lambda\alpha \frac{1}{f}

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

\lambda = \frac{h}{p}= \frac{h}{mv} 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 v(phase)=\frac{\omega }{k}. 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 \frac{\bar{h}}{2}. We can formulate this principle in 2 ways: first, by introducing momentum and position; and second, by introducing energy and time:

\Delta x \Delta p_{x} \geq \frac{\bar{h}}{2} which relates to the form with energy and time, \Delta E \Delta t \geq \frac{\bar{h}}{2}. Given that \bar{h} = \frac{h}{2\pi }.  We can also write the above formulas as follows:

\Delta x \Delta p_{x} \geq \frac{h}{4\pi } and \Delta E \Delta t \geq \frac{h}{4\pi } respectively. 

Example Problems

Problem 1

An electron travels at a speed of 6 x 103 m/s and has an accuracy of 0.003%. Calculate the minimum value for the uncertainty in determining the electron’s position.


(i). First, show the momentum \Delta p_{x} = m\Delta v_{x} = mfv_{x}

(ii). Next, calculate the uncertainty \Delta x \geq \frac{h}{2mfv_{x}}

\frac{1.055 \times 10^{-34}}{2(9.11 \times 10^{-31})(0.003 \times 10^{-2})(6\times 10^{3})} = 0.32 mm.

Problem 2

Calculate the line width \Delta f using the uncertainty principle given that \tau = 1\times 10^{-6} s


(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 \Delta f = \frac{\Delta E}{h}

(iii). Next, we can rearrange the uncertainty principle \Delta E \Delta t \geq \frac{\bar{h}}{2} for the change in energy \Delta E \geq \frac{\bar{h}}{\Delta t. 2} .

(iv). Next, we can substitute \bar{h}=\frac{h}{2\pi } which should give us \Delta f \geq \frac{1}{4\pi \tau }

(v.). Finally, we can substitute the value for \tau given into the above formula for (iv) and will get: \Delta f \geq 8 \times 10^{4} Hz

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