Recent content by neural_jam

Did you mean divide it by the density of Si? If so you get cm, which is very useful, thankyou. However, I don't quite understand what I'm doing when I do this; I mean how do they get to the g/cm2 units in the first place? Also, why do they use this unit for range instead of cm or m or whatever...

Hi, I'm trying to find the range of α-particles in silicon; I can work out the stopping power, \frac{dE}{dx}, but the reference I'm using then states that I need to integrate it as follows: R=\int_T^0 (-\frac{dE}{dx} ) ^{-1} dE The thing is, I'm not sure as to how to even start this...

Ok, that's very helpful, thank you :smile:

Oops, just realize that it's actually \int_0^π\int_0^π\int_0^R 1.22 x 10-2 \frac{A}{4\pi r^2} r2sinθ dr dθ d\phi Since it's 147 Bq m-3 but only 1.22 x 10-2Bq within the actual hemisphere, which takes into account the range of the alphas in air

That's exactly what I'm after (at the moment) thank you! :biggrin: I'm not quite sure what you mean by making 'A' a function of the angles though? I'll more than likely need to make it more precise, so it would be good to know! Incidentally, I used the above equation and got 1.25 x 10-4Bq...

lol yes, which is why I am talking about integrating over the volume of a hemisphere which has the radius of the range of the α-particles in air and centred on the detector. Would it be a valid approach to integrate the 147\frac{A}{4\pi r^2} equation? so it would be...

Hi, firstly I'm not sure if I'm posting in the correct forum, if not could I request that the moderators move it to the right one. Ok, so my problem is pretty much as I stated in the title. The background to this is that I have been asked to roughly calculate the number of alpha...

Hmm... I'm not sure that this is the case; correct me if I'm wrong, but you sound like you're from the US? I'm in the UK, I suspect that companies are a little less litigious than over there, certainly in the area of research. I got the impression that the people interviewing me were...

Thanks for the response. That's just the think, I finished my course almost a year ago now, and I feel like I'm not used to having to think any more, and I wanted some practice questions to keep my brain active, and to get back into the kind of thinking demanded by that particular style of...

Hi all :-) A little while back I went for an interview for a physics based job which, unfortunately, I completely screwed up. Part of the problem was that when they asked me some really basic physics questions, the sort of thing I haven't done since first/second year, I was a.) put off...

Ah, thankyou! Yes, it was a silly mistake, I forgot to change E=2 \times 1.6 \times 10^{-19} to E = 2, yes it works out fine now. Thank you both, I have an exam tomorrow where this will likely be part of a question or two, so I'm very grateful for the help!

Hi, thanks for the reply! Unfortunately, if I put in the factor of c^2, then n^2 becomes ~2.13^{-17}, even further off the 134 that it should be :-(

Ok, so this example question is in my lecture notes: Homework Statement "For a 1-D quantum well, 5nm width, 2eV barrier height, calculate the number of energy levels in the well." Homework Equations So the equation given is a familiar one: E_n=\frac{n^2 \pi^2 \hbar^2}{2 m L^2}...

hmm... that was my initial thought too, but then the wavelength of the wave on the glass rim is then half that of the emitted tone... ok, thanks for your help :-)

ok, that leaves me with a second question; I just did a rough calculation, and it turns out that the wavelength is about the same as the circumference of the glass, which would lead me to expect to observe two nodes and two antinodes as I moved a microphone around the glass, i.e., two...