Recent content by Benlaww

a) V=(4/3)pi(r^3) N=M/m_n (M=mass of neutron star, m_n=mass of neutron) Subbed into E_f = (hbar^2 / 2m) (3(pi^2)N / V)^(2/3). T_F = E_F / k_B b) dU = (dU/dS)_s dS + (dU/dV)_s dV p = -(dU/dV)_s dV V=(4/3)pi(r^3) -> r = cubedroot(3V/4pi) subbed into U_g = -(3/5)(G M^2 / r) take (dU/dV) plug into...

I have completed part a, from which I got the expression: Cv = 3KTn/(T_f) For part b, the first term is the electron contribution and the second term is the phonon contribution. I'm stuck on how to estimate the fermi energy for the potassium metal. I'm thinking I only need to consider the...

Thank you!

I think I got part 1, I've got that Cl-37 + v --> Ar-37 + e- Changing the molecular mass to 0.172kg, I then get 5.68 * 10^28 molecules so then the number of Cl atoms would be four times this? giving 2.27 * 10^29. So the total area is 2.73 * 10^30 fb.

Thank you, I've got that I have 5.89 * 10^28 molecules of C2Cl4 but I'm still a bit confused on how it all relates.

I'm still a bit confused with working this out, I have got the volume of the 'pipe' as 10 m^2 and so the density times volume gives 16220 kg and then I've divide this mass by the mass of 1 Cl atom to give 2.64*10^-26 atoms?..

Would the element it decays to just be an isotope of Cl or am I supposed to be able to tell which new element it would be?

For the first part I thought you'd have either, p + anti-v -> n + e+ and n + v -> p + e-, but I thought it'd probably be the latter as it's a 'normal' neutrino not an anti neutrino? But do I need to include the actual elements in the equation? For the second part I have multiplied the density...

I'm not too sure where to start, do I differentiate the equation?

E = E' + K Ahh I've figured it out now, I got my final wavelength as 1.39*10^-10 m. Thank you!

K= 1/2 (9.11*10^-31)(8.2*^6)^2 (wl)' = hc/k = 6.5 nm (wl)' - (wl) = 6.36 nm cos(theta) = 1- (((6.36nm)(9.11*10^-31)(3*10^8))/(6.63*10^-34)) = -2620.7 theta = error

Thank you so much! This is a great help!

So H.2πr = I _ free enclosed? Does r stay as r or does it go to R?

So would it be ∫ H . 2πr dr with bounds 0 and R?

so dS = r dr dθ Z as there's no terms in θ, after canceling terms becomes: (j_0/a^2) ∫ sin(ar) dr - (j_0/a) ∫ rcos(ar) dr