Recent content by Max Eilerson

All correct except the 5p -> 3d, as l changes by 2. I knew that transitions needed change in l =+- 1 but thought the 5p and 3d states were interchangeable?

An electron is excited to the 5p state in potassium, list all the possible dipole transitiosn that can occur as the electron relaxes to the 4s level. (Ignoring fine structure effects) I've got 5p -> 5s -> 4p -> 4s 5p -> 4s I'm not sure whether to also include 5p - > 3d -> 5s...

Homework Statement The L shell absorption edge of rubidium is at 1.81kev and the K (n = 2 -> 1 ) emission line is at 13.41 kev. Calculate the energy of the K-Shell electron in rubidium, and hence estimate the effective nuclear charge for K-Shell electrons. Homework Equations Alkali so E =...

I don't include units when I'm working with variables ( I keep track of them when dealing with long derivations of course), but I always put them in when I write the numbers down at the end. I think people who don't should be penalised even if they get the right units for the final answer.

http://www.edexcel.org.uk/quals/gce/chemistry :).

In a report should I be quoting uncertainties as 2.435(36) m or 2.435 +- 0.036 m. It seems the first way is usually used when quoting a standard deviation error and the second when quoting a resolution error.

Couldn't you do it on a computer game such as Project Gotham Racing 3? Perhaps look at the modelling they use.

I assume it means λmax = 1 Angstrom = 0.1nm. If you're just given a wavelength you can only use wien's displacement law.

If I'm calibrating a magnet between 0.5T and -0.5T. Should I go to like 0.1, and flip the current to -0.1 then back to 0.1, onto 0.2, -0.2 etc. Or go in increments from 0T to 0.5 T, then 0T to -0.5T, after putting the magnet through a hysteris loop? I'd imagine the first way, but am wondering...

Huge energies (accelerations) are required to produce these particles. You might want to look at the rest mass of some of the fundamental particles

1 = \int \rho(r) d^3r

Just found this page which says normalised means you can omit the e http://www.physics.rutgers.edu/ugrad/418/FormFactors.pdf (page 3) The series expansions give me one in that limit with the e omitted:).

So I evaulated the integral with \rho(r) = \frac{3e}{4 \pi R^3} q = momentum transfer, e = proton charge. \frac{3e\hbar^2(\hbar\sin[\frac{qr}{\hbar}] - qr\cos[\frac{qr}{\hbar}])}{q^3 R^3} + C Evaluated between R and 0, (F(q) = 0 between R and infinity since p(r) = 0.) F(q) =...

I suspect that your getting confused with signs. You do know how to cross multiiply right? \frac{1}{d_i} - \frac{1}{30 - d_i} = \frac{1}{f}

Of course you could just think how far from a lens a focused image is :).