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Degree: Theoretical Physics (Masters)  Birmingham University
I am a third year Theoretical Physics student at the University of Birmingham. I am passionate about all things physics and am particularly keen on anything with a strong mathematical element. I am very patient and have a real interest in teaching and the communication of science in general. During my time in the Air Cadets, I designed and delivered a teaching programme for teenagers of varying ages and abilities in order to teach subjects related to the principle of flight, history, navigation amongst other subjects.
Tutorial Sessions
I believe that everyone has their own unique learning style, particularly when it comes to subjects like maths and science. For this reason I feel it is important that you get to choose the pace and style of the sessions. If, however, you feel unsure of where to start then we can begin with a general introduction to the area in question, slowly building up your confidence before we move onto questions, exam technique and problem solving skills.
As a rule, I think its important to enjoy the subjects you study in order to achieve in them. I hope that through a variety of teaching approaches and tasks it will be possible to improve not only your knowledge of a subject but your motivation and interest as well!
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For any more information on the sessions or anything at all, don't hesitate to send me a message or book a 'Meet the Tutor' session.
I look forward to meeting you soon!
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24/06/2014 
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Modulusargument form implies that we should express z in terms of its straight line distance from the origin and the angle this straight line would make with the x axis. This is expressed as z = r e^{i*theta} where r is the modulus and theta is the argument. We thus wish to express r and theta in terms of x and y.
This problem is best solved visually by considering an Argand diagram for the general complex number, z. In this way we can see that z is represented as a point x units along the xaxis and y units along the yaxis, forming a right angled triangle with the vertical and horizontal.
As with any right angled triangle, using Pythagoras, we can see that the length of the hypotenuse (I.e. the distance from the origin, the modulus) is given by r = sqrt(x^{2}+y^{2}). Similarly, using basic trigonometry we can also see that the angle between this line and the x axis (theta, the argument) is given by theta = arctan(y/x).
This means that in modulusargument form:
z = sqrt(x^{2}+y^{2}) exp(i*arctan(y/x))
see moreWhile this question may appear unfamiliar at first, it is just a quadratic equation in disguise. Rearranging such that it equals 0:
x^2 + 5x  36 = 0
Now this equation is ready to be solved! We can approach this in a couple of ways. I'll go through the factorisation method and using the quadratic formula.
Factorisation Method
This method is often the quickest way but requires a bit of practise to be able to spot the solution. If we look at a general factorisation where the coefficient of x is 1 in both brackets, i.e.
(x+a)(x+b) = 0
We can see when we expand this, we get
x^2 + ax + bx + ab = 0
x^2 + (a+b)x + ab = 0
Going back to our example, this shows us that we need to think of numbers such that a + b = 5 and ab = 36. The tricky part is actually thinking of these numbers.
Because we know that ab is a negative number, we know either a or b must be negative because a negative times a positive is a negative. We now need to trial and error a few potential numbers to find our answers. With a bit of good guessing we can see that a = 4 and b = 9 will work because 4x9 = 36 and 4+9 = 5.
This gives us
(x4)(x+9)=0
For this to equal zero, one of these brackets must equal 0. If the first one equals 0 then,
x4 = 0
x=4
or the second one,
x+9=0
x=9
So our answer is x=4 or x=9.
If you found it difficult to guess the right numbers then keep practising but, in the meantime, the formula method always works without guessing but does take a little longer...
Formula Method
For a general quadration equation ax^2 + bx + c = 0, the quadratic formula is given by
x = (b +/ sqrt(b^2  4ac))/2a
For our example, a = 1, b = 5 and c = 36, so:
x = (5 +/ sqrt(25 + (4x36)))/2
x = (5 +/ sqrt(25+144))/2
x = (5 +/ sqrt(169))/2
x = (5 +/ 13)/2
Taking the + option:
x = (5 + 13)/2
x = 8/2
x = 4
Taking the  option:
x = (513)/2
x = 18/2
x = 9
Therefore, x=4 or x=9 as expected from our first method.
see moreThe photoelectric effect is the emission of electrons from a metal when light is incident on the surface. The effect is due to the absorption of energy by the surface electrons causing them to become excited and thus released from the metal surface.
The wave theory for light would suggest then that, as long as we were to wait long enough, light source of any frequency should eventually cause the emission of electrons. It would also suggest that the kinetic energy of the emitted electrons would be controlled by the intensity of the light because, you would think, the brighter the light, the more energy it is providing.
However, it was observed that light intensity had no impact on the kinetic energy of the electrons and also that lower frequency light would not emit any electrons regardless of how long we wait.
On the other hand, the particle theory of light suggests that light is transmitted in 'packets' of energy called photons, each of energy E=hf, where f is the frequency and h is Planck's constant. It also suggests that electrons would absorb these photons on a oneonone basis. This theory perfectly describes the observation as it suggests that unless an individual photon has enough energy, an electron will not be released from a metal. It also means that it doesn't matter how long we wait, if the photon doesn't have enough energy, the electrons will never be emitted. Since photon energy was only dependent on frequency, this would also explain why the electron kinetic energy was dependent on frequency and not the light intensity.
Other examples include the instantaneous emission of electrons and considerations of the work function of the material.
see more