Recent content by Noirchat

Great I'm on the right track then. Hmm I'm not sure why they're incorrect. Is it to do with the Compton constant?

I'd really like some explanations please, just looking at part (a) at the moment :) Homework Statement Suppose three 1.02 MeV X-ray photons are Compton scattered by single collisions with nearly stationary electrons. Photon 1 is scattered backwards, in the direction opposite to its original...

No one? :-(

Would like to see if I've understood this question somewhat. :-) Homework Statement A doubly ionised Lithium atom (Li^++) has two of its normal complement of electrons removed. The energy levels of the remaining single electron ion are closely related to those of the neutral Hydrogen...

Ok, that makes sense to me. So have i used the wrong equation?

Isn't it the minimum energy needed to dislodge an electron so it can move between energy states?

Homework Statement In a set of experiments on a hypothetical one-electron atm, you measure the wavelengths of photons emitted as electrons return to the ground state (n=1), as shown in the energy level diagram. You also observe that it takes 17.50 eV to ionise this atom. Diagram shows...

I assume the graph will look like this: http://media.smashingmagazine.com/wp-content/uploads/2011/09/sine.png. ?

Well since the numeral/letter in front of 'sin' gives you the range of the graph wouldn't it be: y = Asinx?

Got the frequency from another q mixed up. Again; (a) λ=v/f λ = 3 x 10^8/109000000 = 2.7523 Momentum of photon = 6.626 x 10^-34/2.7523 =2.407 x 10 ^-34 (b) momentum of electron = mv v = p/m p =2.407 x 10 ^-34 m = 9.10938 x 10^-31 v = 2.642 x 10^-66 m/s ///...

Thank you! :) So for part (a), λ=v/f λ = 3 x 10^8/10490000 λ = 28.599 Momentum of photon = 6.626 x 10^-34/28.599 =2.32 x 10^-35 (b) momentum of electron = mv v = p/m where p = 2.32 x 10^-35 m = 9.10938 x 10^-31 v = 2.5468 x 10^-5 That is very true, i don't want...

I was thinking about m/s but nevermind that! Does it represent half a 'normal sine wave'? I'm actually learning quite a bit, so sorry if this drives you a bit crazy... is it right to say for part (i) that half a wave = 33cm. And therefore a full wave = (which equals one wavelength) =...

Thanks for your reply guys. I actually didn't know c=fλ only applied to electromagnetic waves! And wasn't sure what the nodes/anti-nodes did. I have tried answering the questions again: i) v=fλ , the question has given me 33cm which I've converted to 0.33 and it's given me 264 Hz. λ =...

Hi guys, i'd really like to understand physics but for some reason I'm not coping. Not sure if it's the calculations or something. But I'm willing to learn, I'd appreciate some help with this problem:Homework Statement A B string on a guitar with a length of 33cm is held fixed at both ends...