Confusion regarding Quantum numbers

[subhradeep mahata]

Homework Statement

Given in the "Attempt at a solution section".

Relevant Equations

Given in the "Attempt at a solution section".

Problem Statement: Given in the "Attempt at a solution section".
Relevant Equations: Given in the "Attempt at a solution section".

Problem Statement: Given in the "Attempt at a solution section".
Relevant Equations: Given in the "Attempt at a solution section".

I am having some serious confusion regarding the following terms:
1) According to Bohr's quantization of angular momentum, mvr=nh/2π. My question is what kind of angular momentum is that? Is the the "orbit angular momentum"?
2) Orbital angular momentum= {l(l+1)}^1/2 BM. Again, I fail to understand what it's significance is.
3) I have been told that the spin angular momentum = {S(S+1)}^1/2 BM where S=n/2. However, I also read somewhere that the value of S is fixed for objects spinning around their axis (1/2 for an electron, +-1/2 is the spin quantum number). So, which of them is correct?
4) Whenever we say "magnetic moment", do we mean the "spin only" magnetic moment which is equal to {n(n+2)}^1/2 BM?
5) This one is not related to quantum numbers. We know that de broglie wavelength for an unchargen particle in non-relativistic situation=h/p where p is the momentum (=m*v). For gaseous molecules, we take 'v' as the root mean square speed (={3RT/M}^1/2). My question is, in the formula or root mean square speed, 'M' is the molar mass of the gas, so to calculate the momentum (=m*v_rms), do we have to take the molar mass of the gas in place of 'm'?

Please have a look at them and reply as soon as possible. Thanks in advance.

 

[mjc123]

1) This is orbital angular momentum in the Bohr model, but it's not what is meant by "orbital angular momentum" in QM, which depends on l, not n.
2) One way to look at it is like this: For an electron with quantum numbers n, l, the wavefunction has a total of (n-1) nodes, of which l are angular nodes. The more nodes, the higher the energy, and the more angular nodes, the higher the orbital kinetic energy (and also angular momentum). Note that an s orbital has zero orbital angular momentum, in contradiction to the Bohr model.
3) Who told you S = n/2? I never heard that before. You are right that S = 1/2 for an electron, m_S = ±1/2. (Note: S = n/2 if n is the number of parallel unpaired electrons - nothing to do with the quantum number n!)
4) I don't understand this, as spin is nothing to do with n. (Note: if n is unpaired electrons again, {S(S+1)}^½ = {n(n+2)}^½/2)
5) For the momentum of a particle, you need the mass of the particle, not the molar mass. You can also use the particle mass in the expression for rms speed, if you write it as {3k_BT/m}^½.