Black Body Radiation

A black body is an object which absorbs all the radiation falling on it. The best way to imagine a black body is to imagine a hollow ball, coated matt black on the inside with a tiny hole in it. All the incident radiation going through the hole is absorbed and is unlikely to be re-remitted out of the hole. A black body in thermal equilibrium (that is, at a constant temperature) emits electromagnetic radiation called black-body radiation. The radiation emitted has a spectrum that is determined by the temperature alone.

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This link is helpful. It shows the shape of the emitted radiation curves at different temperatures which can be changed by moving the sidebar. 

Doubling the temperature increases the area underneath by a factor of 16, because of the Stefan-Boltzmann Law, which  states that the total energy P radiated per unit surface area of a black body across all wavelengths per unit time is proportional to the fourth power of Kelvin temperature.  The constant is the Stefan-Boltzmann constant.

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Notice how the maximum wavelength shortens as temperature increases. An object at 5000K will tend to glow red, at 6000K it will be much bluer.

The area under the curves is a measure of the energy of the spectrum at a particular temperature. Doubling the temperature multiplies the energy emitted by a factor of 16 ( 2 to the fourth power)

 

 

 

 

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Nuclear Fission

This post is a description of how neutrons produced in a fission reaction may be used to initiate further fission reactions – a chain reaction.  Both power stations and weapons use a variety of nuclear ‘fuels’ – poorly named since a fuel is more usually a descriptor for a substance which burns and combustion is exothermic, releasing heat. A common nuclear fuel is uranium 235, which is the only naturally occurring fissile material.

We should know that only low-energy neutrons (< 0.2 eV) favour nuclear fission.

Principle.

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Unstable nuclei capture a slow neutron of low energy. Why are the neutrons ‘slow’? Because the U nucleus is said to have a large ‘cross section’ for slow neutrons – rather like using a big, oversized racket to hit a small ball – the probability of the ball being hit is greater. The U spontaneously divides into 2 smaller parts, approximately equal in size plus 2,3 or 4 more neutrons which may be captured by other nuclei, producing a chain reaction (see diagram). Unless this is controlled, if enough fuel is present (a 7kg ball is enough – the critical mass), the reaction goes out of control and basically becomes explosive. If the ball is too small, neutrons will tend to pass through it, we need enough uranium present to ensure that the probability of neutron capture is high.

The mass of the fragments is less than the mass of the original parts, thus mass ‘loss’ and energy ‘gain’ are equivalent. [E=mc2].

Screen Shot 2015-05-13 at 16.13.30In a reactor for power generation, the KE of the moving fission products is converted into heat when passed through water in a very short time. Sometimes, we’re asked to find the kinetic energy of the fragments using equivalence of thermal and kinetic energy.

Enrichment.

is the process by which the percentage of fissile uranium is increased. Uranium as mined is about 0.7% by mass. Enriching by isotope separation, gaseous or thermal diffusion and centrifuging boosts this percentage to between 3 and 4% for reactor grade uranium and if the process continues, high purity weapons grade uranium requires almost 90% purity. A crude, inefficient weapon can be manufactured with  a source of 20% or more, however.

Control Rods and Moderators

The control rods absorb the neutrons which keeps the reaction rate relatively constant (rather than letting it grow exponentially). They create a situation where roughly one neutron per fission goes on to cause another fission event. Silver, indium, cadmium or boron are commonly used to make them. In effect, the absorption process creates a new isotope;

Moderators slow down the neutrons without absorbing them. Fast neutrons are more inclined to bounce/deflect off of the surface of a nucleus so slower neutrons actually lead to a greater number of successful fissions i.e. moderators don’t slow the reaction down, they just help it to take place. Commonly-used moderators include regular (light) water (in 74.8% of the world’s reactors), solid graphite (20% of reactors) and deuterium oxide or heavy water (5% of reactors). Their job is to be introduced in controlled amounts into the nuclear pile (bundled fuel rods) to slow down neutrons and so increase the probability of a fission event.

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Answer the Question IB

John Vagabond's Physics and Chemistry Blog

In the IB papers, you will be asked to answer questions using a particular form of words. ‘Describe’ isn’t the same as ‘explain’, for example. Here’s a list – the so-called ‘Objective 3 in the syllabus, plus a translation. Even people who know what they are doing fail to score because they’ve fallen into the trap of misunderstanding the question’s precise requirements.

This screenshot should help. Remember, in paper 2 (HL) question 1 is always data analysis, often with a discussion of errors (see earlier post for more help). Please read and comment.

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Accuracy and Precision

IB exam questions sometimes ask you to think about the difference between accuracy and precision of a set of measurements. Ideally a measurement device is both accurate and precise, with measurements all close to and tightly clustered around the known value. The accuracy and precision of a measurement process is usually established by repeatedly measuring some traceable reference standard. Such standards are defined in the SI system – the whole community knows and uses the same standard values for metres, seconds and so on.

The accuracy of a system of measurement is the degree of closeness of repeated measurements of a particular quantity to that quantity’s actual (true) value.

The precision of a measurement system, also called its repeatability is the degree to which repeated measurements under unchanged conditions show the same result.

We should also incorporate the idea of ‘trueness’. Trueness refers to the closeness of the mean of the measurement results to the actual (true) value and precision refers to the closeness of agreement within individual results. Therefore, the term “accuracy” refers to both trueness and precision.

A couple of diagrams might be helpful On the y-axis we might write “increasing number of measurements”Screen Shot 2013-10-14 at 4.51.27 PM Screen Shot 2013-10-14 at 5.00.48 PM Screen Shot 2013-10-14 at 5.01.18 PM

The last is perhaps the easiest to understand, using as it does the analogy of a target in a shooting range. The first diagram shows good trueness because the average of all the results places it closest to the bullseye. The range or scatter of the shots is large, however, so the accuracy is poor, as is the precision since the shots are widely spaced.

Conversely, the right hand target shows low accuracy and trueness because the mean of all the shots is a long way from the bullseyse, but the shots are nicely clustered together suggesting good precision – the mean of the shots is close to all of the attempts.

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Uniform Electric Fields

If you can do gravity, you can do electrostatics. They are both inverse square laws for non-uniform fields and behave similarly for uniform fields too.

Remember this diagram? This time, imagine two parallel metal plates – very large, separated by a few mm in air. A primitive capacitor, in other words. The top plate is positively charged by connecting it to the + terminal of a battery and electrons from earth will be attracted to the bottom plate. Because the charges are all of the same kind on each plate and the surfaces are flat, they spread out as far as they can, like a carpet. Now, imagine a charge of  say +3mC (in blue) somewhere in between the plates. It’s experiencing a downward force, repelled by the top plate and attracted to the bottom one. This is quite a large blob of charge, since an electron has a charge of 1.6 x 10-19C.

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Why is the force always vertically down (or up, depending on polarity)?

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If we imagine our carpet of charges, and the charged particle in between them, each + charge exerts a repulsive F on the test charge in the middle. Each – charge on the bottom exerts an attractive F along the same line, as the diagram shows. Each + is paired with a – in other words. Taking two at random the resultant F is seen to always act downwards, and is the same size, wherever we are in between the plates. There is a uniform electric field between the plates in the same direction as the force on the charge. An electric field is a region where charges experience an electric force. Numerically, it is the force per unit charge, so we can write:

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E is a vector because F is a vector. We now see that the red lines on the diagram show the direction of E. They are uniformly spaced since the field is uniform, in other words, the field is everywhere the same.

Let’s put some numbers on our original diagram. The dotted lines are lines of equal potential, or equipotential lines (or surfaces). We have to do electrical work on a + charge to climb up from one equipotential to the next, just like we have to do mechanical work against a gravitational field.

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The p.d between each equipotential is 5V in each case and let’s suppose that the plates are 6mm apart ( not shown on the diagram).

Now, watch this. We know that a volt is a joule per coulomb. Also, (electrical) work done = force in N x distance in m. So, we can write that:

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So, the uniform electric field between two parallel plates can be written down as:

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In our example above, and being careful about units, E = V/d = 15/0.006 = 2500V/m. On a charge of 3mC, the electric force F between the plates is Eq = 2500 x 0.003 = 7.5N

Edge effects. IB asks this, A level doesn’t. At the edges of the plates there are no more partners for the charge in the middle to pair up with , so the field lines curl outwards. Be careful, if you’re asked for this diagram, it has to be tidy, with clear, non-overlapping, non-dotted lines WITH ARROWS, starting and finishing at a plate, like this.

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Gravitational Fields 2. Non-uniform (radial) Fields

This post follows on from this earlier one which introduced Newton’s Universal Law of Gravitation.

An object with mass creates a gravitational field around it sucking mass towards it – like a whirlpool. Imagine a paper boat. It will always be pulled towards the centre of the hole.Screen Shot 2013-09-12 at 4.05.38 PM

The gravitational field strength (vector) g is defined as the gravitational force at a given point divided by the mass of an object at that point so g=F/m

On earth g = 9.81Nkg-1 (acc due to gravity). Slightly variable, equatorial r >polar r, thus g greater at the poles, neglecting the spin of the earth which will tend to reduce effective value.

g = constant near the Earth’s surface, falls as as we move away.

Gravitational field lines (parallel) are perpendicular to equipotential surfaces (zero change in GPE  when moving around on them, i.e, the field is conservative. The same work is done against the gravitational field irrespective of the path taken, in other words we have to do work to climb the gravitational hill, but it doesn’t matter which path we take – the same amount of work is done. The spherical earth is an equipotential surface.  The Space Station orbiting the same distance above the Earth is another. If we were to plot EQUAL CHANGES in potential as a series of concentric circles, would they be equally far apart, or not? Answer: no they wouldn’t. Lines in regions of higher field strength would be closer together. (fave exam q)

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How does g vary with distance from the Earth? You could try making a spreadsheet like this. You could try plotting F (from Newton’s Law of Gravitation) against r for various distances from the centre of the Earth, beginning with r = earth’s radius of 6400 km. For masses, just use the mass of the earth and a 1kg test mass which is what I did. Shows the inverse square relationship nicely.Screen Shot 2013-09-12 at 3.58.45 PM

Screen Shot 2013-09-12 at 3.18.13 PMLines of force. Check out this link and play with the animation. Notice the arrows’ length (closer at small r)  and direction (attractive F so always towards the centre) The Earth’s field is radial, like the spokes of a wheel.

This is the superposition of the fields due to the Earth and the Moon.Screen Shot 2013-09-12 at 3.40.24 PM

There’s a place between the Earth and the Moon where the gravitational forces cancel out. Here’s a ratio problem. Neglecting the gravitational effects of other bodies like the Sun, imagine a 1kg mass somewhere along a line between the Earth and the Moon. How far from the Earth would it be when the gravitational forces of the Earth and the Moon balance out?

(mmoon = 7.35 x 1022 kg, Earth/Moon distance = 3.8 x 105 km)

{Hint: if earth moon distance = 1 unit; at a distance r from the earth where g=0, distance from moon=(1-r). Not easy but try using Newton’s Law of Gravitation}

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Equipotential surfaces (gravitational)

Near the surface of the earth, we represent the gravitational field acting on an object by an arrow acting downwards, representing attraction to the centre of the Earth (brown lines). The field lines are straight and parallel, as shown. The field is (or can be thought of as) uniform, or g is invariant with h.

If acting on a mass m, this, clearly, is the weight of the object (blue)

Horizontal lines represent surfaces of equal potential, in other words, bodies of mass 1kg having the same gravitational potential energy. No energy is transformed when a body moves over an equipotential surface. In our diagram, if the mass is moved horizontally, no work is done by it or on it. Moving it vertically upwards requires us to do work on it.Screen Shot 2013-09-12 at 2.40.49 PM

Evenly spaced equipotentials in a uniform field such as that close to the Earth are perpendicular to the field lines. Let g=10Nkg-1 and let’s imagine we, a mass of 60kg, are walking up a hill 30m high. Each ‘step’ is 10m apart.

Screen Shot 2013-09-12 at 2.40.59 PM

For every 10m gained, the 60kg mass acquires 6000J of energy (60kg x 10N/kg x 10m)

Increase in gravitational potential is 10J/kg for every metre of height gained. The gravitational potential difference per meter is 10J/kg m-1

How much energy would be required to move double this mass through a gravitational potential difference of 1.2kJ/kg?

If the field was that on the Moon, one-sixth of its value on Earth – assumed uniform, what height would you have to raise it to above the lunar surface?

When g isn’t constant, the field is no longer uniform. More later.

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