Physics Lens

18 July

20. Nuclear Physics

Seng Kwang Tan Subject Content

The Nucleus

existence and size demonstrated using the Rutherford $α$ -scattering experiment.
consists of nucleons (protons and neutrons)
isotopes of an element share the same number of protons but different number of neutrons.

Nuclear Reactions

nuclear reactions involve two or more reactants.
represented using the form: $_{7}^{14} N +_{2}^{4} H e \to_{8}^{17} O +_{1}^{1} H$
for a reaction that releases energy, mass-energy of reactants = mass-energy of products + E,
where $E = m c^{2}$ and m is the mass defect (difference in mass between the products and reactants).
binding energy is the energy released when the nucleus is formed from its separate protons and neutrons. The same amount of energy is required to break up a nucleus into its constituent nucleons.

binding energy per nucleon ( $\frac{B . E .}{A}$ ) is an indication of the stability of a nucleus, where B.E .is binding energy and A is the nucleon number. You need to know how to sketch its variation with nucleon number. (The following video explains the shape of the $\frac{B . E .}{A}$ versus A graph and why it peaks at $^{56} F e$ .

nuclear fission is the disintegration of a heavy nucleus into two lighter nuclei of comparable mass with the emission of neutrons and release of energy.
e.g. $_{92}^{235} U +_{0}^{1} n \to_{92}^{236} U \to_{56}^{144} B a +_{36}^{90} K r + 2_{0}^{1} n + E n e r g y$
nuclear fusion occurs when two light nuclei combine to form a single more massive nucleus, leading to the release of energy.
e.g. $_{1}^{2} H +_{1}^{3} H \to_{2}^{4} H e +_{0}^{1} n + E n e r g y$

The following quantities are always conserved:
- proton number & neutron number
- momentum
- mass-energy

Radioactive Decay

spontaneous and random emission of radiation from a radioactive nucleus.
- $α$ particle – helium nucleus
- $β$ particle – electron
- $γ$ particle – electromagnetic radiation

http://youtu.be/Qlb5Z8QBpcI

$A = - \frac{d N}{d t} = λ N$
where A is the rate of disintegration or activity, N is the number of radioactive nuclei and $λ$ is the decay constant.
$x = x_{0} e^{- λ t}$
where x could represent the activity, number of undecayed particles or received count rate.
half-life ( $t_{\frac{1}{2}}$ ) is the average time taken for half the original number of radioactive nuclei to decay.
From $x = x_{0} e^{- λ t}$ ,
$\frac{x}{x_{0}} = \frac{1}{2} = e^{- λ t_{\frac{1}{2}}}$
$\Rightarrow - l n 2 = - λ t_{\frac{1}{2}}$
$\Rightarrow t_{\frac{1}{2}} = \frac{l n 2}{λ}$
You may also use $\frac{x}{x_{0}} = {\frac{1}{2}}^{\frac{t}{t_{1 / 2}}}$ , as shown in the following video.

13 June

P-N Junction

Seng Kwang Tan Teaching Resources

The following is the transcript for a video that I will be making to explain how a P-N junction works.

A p-n junction, as the name suggests, is the boundary between two types of semiconductors: P-type and N-type.

For an intrinsic or pure semiconductor such as silicon which has 4 valence electrons, each atom is bond to 4 other neighbouring atoms.
The p-type semiconductor is one with excess holes due to the addition of dopants to intrinsic semiconductors. Elements such as boron or phosphorus from Group III of the periodic table all contain three valence electrons, causing them to function as acceptors when used to dope silicon. When an acceptor atom replaces a silicon atom in the crystal, a vacant state ( an electron “hole”) is created, which can move around the lattice and functions as a charge carrier.

The n-type semiconductor is one doped with Group V elements which have five valence electrons, allowing them to act as a donor; substitution of these atoms for silicon creates an extra free electron. Therefore, a silicon crystal doped with boron creates a p-type semiconductor whereas one doped with phosphorus results in an n-type material.

When the two types of semiconductors are put together, electrons diffuse across the boundary to combine with holes, creating a depletion region where there are no charge carriers. An electric field is also set up in the depletion region because the group III atoms are now negatively charged, having gained one more electron and the group V atoms are now positively charged, having each lost an electron. This electric field prevents further charges from diffusing across the boundary.

That is, until a potential difference is applied. The p-n junction serves now as a diode. We shall illustrate this with a single cell attached to the device. In the reverse-biased mode, the positive terminal is connected to the n-type semiconductor while the negative terminal is connected to the p-type end. This causes more electrons to move away from the depletion region in the n-type semiconductor and for more holes to be filled in the p-type semiconductor. The result is a widened depletion region and a larger opposing electric field.

In the forward-biased mode, the positive terminal of the cell is connected to the p-type semiconductor while the negative terminal is connected to the n-type end. The potential difference provided offers the electrons in the n-type semiconductor a push to overcome the small electric field formed across the depletion region and flow across to the p-type semiconductor which it then passes from hole to hole into the positive terminal.

12 June

Planning Question

Seng Kwang Tan 21 Practical / Planning

The following is the script that I used for recording the video above for our e-learning day. I’m posting it here as I can edit it easily via wordpress’s mobile app, and because I have LaTeX enabled here.

Aim:

The aim of this planning question is to investigate how the resonant frequency of a wire vibrating in its fundamental mode depends on the tension in the wire.

The independent variable is the tension in the string which can be varied by hanging masses at one end of the wire and dangling that end over the edge of a table on a pulley. The tension is represented by the symbol.

The dependent variable is the resonant frequency of the fundamental mode in the wire. The fundamental mode consists of two nodes at both ends of the wire and one antinode in the middle.

In other words, in our experiment, we shall vary tension and measure the resonant frequency of the fundamental mode.

The length of wire between the supports is kept constant throughout the experiment and we shall use the same wire throughout so that the mass per unit length is kept constant.

Procedure:

The experiment will be set up according to this diagram. One end of the wire is first tied to a fixed object. The other end is hanging over a pulley clamped on the edge of the laboratory table and tied to a mass hanger. To control the length of the wire, place a bridge at each end. Only the length of the wire between the two bridges will be vibrating. We shall keep this length constant for the whole experiment.

Record the mass of the mass hanger m and determine its weight. The weight mg is taken to be equal to the tension T on the string.

A uniform horizontal magnetic field is generated by a pair of large electromagnetic coils on both sides of the wire. The wire is connected on both ends to a function generator. An alternating current is produced by the function generator and its frequency can be varied using the same apparatus. The function generator should have a display that enables us to read the frequency.

The resulting magnetic force acting on the wire will be driving the oscillation of the wire at the frequency shown on the function generator. By adjusting the frequency of the alternating current until the fundamental mode of a standing wave is formed on the wire, we can record the resonant frequency f_ofor the corresponding tension T.

We shall repeat the experiment for different values of m by adding known amounts of mass (e.g. 50 g increments each time) onto the mass hanger. All the values for the mass m and resonant frequency for the fundamental modes f_o should be recorded and tabulated.

The tension in the wire is then calculated using the equation T = mg.

Assume that the resonant frequency for the fundamental mode f_o and the tension T follows the equation f_o = kTⁿ where k and n are constants. Then lg f_o = lg k + n lg T. Plotting a graph of lg f_o versus lg T, we can conclude that the assumption is correct if a linear relationship is observed and we can obtain the values of n and k from the gradient and the vertical intercept of the graph respectively.

As a safety consideration, the person conducting the experiment should wear goggles as the wire at high tension might suddenly snap or come loose. Always handle power supply with care.

As a precaution to improve reliabiliy, we can place a white card behind the vibrating wire so it can be seen easily. To make sure that the weight of each of the slotted masses is as indicated, measure them on a weighing balance. FInally, ensure that the pulley is smooth by measuring the tension in the wire using a force meter to check that it is indeed equal to the weight of the slotted mass.

01 June

Boyle’s Law

Seng Kwang Tan 08 Temperature and Ideal Gases, Demonstrations

Using a hand-operated vacuum pump, we can demonstrate the relationship between pressure and volume of a gas. According to Boyle’s law, the pressure of a gas of constant mass and temperature will be inversely proportional to its volume.

In our demonstration, we will reduce the ambient pressure within the sealed container, hence allowing the higher internal pressure of a balloon to cause it to expand. When the volume within the balloon increases, the internal pressure can be observed to decrease until it is in equilibrium with the surrounding pressure.

While the relationship between pressure and volume is not exactly obeying Boyle’s law due to additional factors such as the tension due to the elastic property of the balloon, it does demonstrate an inverse relationship.

19 May

Polarization with 3 Filters

Seng Kwang Tan 11 Wave Motion, Demonstrations 0 Comments

In what seems like a counter-intuitive demonstration, we can place a polarizing filter in between two other filters which do not transmit light in order to cause light to pass through again.

This is because each filter will permit the components of electric field vectors of the electromagnetic waves that are parallel to its axis of polarization according to the equation $A = A_{o} \cos θ$ where $A_{o}$ is the original amplitude of the unpolarized wave incident on the filter and $θ$ is the angle between the electric field vector and the axis of polarization. Each time the wave passes through a filter, it undergoes a reduction in amplitude according to the equation so that by the third filter, its resultant amplitude is
$A = A_{o} \cos θ_{1} \cos θ_{2}$
where $θ_{i}$ is the angle between the axis of polarization of the i^th filter and the electric field vector direction of the incident light on the i^th filter.

According to Malus’ law, the intensity of the light that passes through these two filters is given by

$I = I_{o} \cos^{2} θ$

$where$ I₀ is the initial intensity and θ is the angle between the light’s initial polarization direction and the axis of the polarizer.

The resulting intensity for light that passes through 3 filters is given by

$I = I_{o} \cos^{2} θ_{1} \cos^{2} θ_{2}$

where $θ_{1}$ is the angle between the axes of the first and second filters and $t h e t a_{2}$ is the angle between the axes of the second and third filters.

18 May

Polarization Using Sunglasses and a Computer Screen

Seng Kwang Tan 11 Wave Motion, Demonstrations 0 Comments

Using a pair of polarizing sunglasses, you can demonstrate the effects of polarization together with a computer screen which is also polarizing. When the axes of polarization of the two polarizing screens are rotated, the brightness alternates between bright and dark.

Light coming from a computer screen is usually polarized. In the video below, when polarized light passes through another polarizer, the intensity of the light is given by Malus’ law:

$I = I_{o} c o s^{2} θ$

where $θ$ is the angle between the two axes of polarization and $I_{o}$ is the original intensity of the unpolarized light.

Only the components of electric field vectors in electromagnetic radiation that are parallel to the axis of polarization of a polarizing filter will be permitted through. Those electric field components that are perpendicular to the polarization axis are blocked by the filter.

Hence, the amplitude of a vector A that passes through is given by $A = A_{o} \cos θ$ . Since intensity is proportional to the square of amplitude ( $I \propto A^{2}$ ), we have Malus’ law.

The purpose of having polarizing filters in sunglasses and computer screens is to cut out glare due to light from other sources.

Physics Lens

20. Nuclear Physics

The Nucleus

Nuclear Reactions

Radioactive Decay

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P-N Junction

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Planning Question

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Boyle’s Law

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Polarization with 3 Filters

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Polarization Using Sunglasses and a Computer Screen

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The Nucleus

Nuclear Reactions

Radioactive Decay

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