Recent content by Kynsuo

Thanks! This is what I was missing. Using the ##T##, ##G## and ##M##, I can find and expression for ##R##, the radius of orbit. Then once I have the orbit I can find the angular diameter. Thanks.

Thanks DocAI. I am given the distance to the centre of the galaxy which is ##d##, I'm assuming that this is the radius of the orbit. I'm confused what this has to do with the period of orbit. Also, don't I need the radius of the black hole to take the ratio of the radius of the black hole to the...

I am confused because the question implies that I need to do some sort of calculation with Kepler's law. I got ##r+d = \sqrt[3]{\frac{T^2 GM}{4 \pi^2} } ## But don't understand why I need this, since I already have the distance and the angular diameter should be ##\arctan (2R/d)## I think I...

High Perok: I think I understand. Basically, the energy must be a constant that doesn't depend on ##r##. Since we have 2 terms of the form ## \frac cr, \; c \in \mathbb C ##, their sum must equate to 0. This also makes sense because the expression gives the first energy state for the H atom when...

Hi PeroK, thank you so much for getting back so fast. This might sound stupid, but are we allowed to do this? Can we 'chose a value for ##a_0## such that the expression is equal to zero' only because this equation is true for all values of ##r##, IE this is not an equation, but an identity?

Hi! Sorry for taking so long to get back. I understand PeroK's logic better now. I alsorealised that there is a typo in my equation (all the ##\hbar##'s are actually ##\hbar^2##) . Thank you for that. The only thing I'm missing is how PeroK went from ##\frac{\hbar^2}{2mr} \frac{2a_0-r}{a_0^2}...

I haven't studied the Hamiltonian in much detail as this came from an introductory foundation course for quantum physics, which I will study in detail next year (all I know is that it is an operator that returns the total energy of a system). What I mean by understand is that I am not able to...

Hi peroK, this is really helpful, and I am confident I can solve this equation for ##a_0##. I think that my problem is that I don't understand why the kinetic energy of the electron in the hydrogen atom is: $$ \frac {\hbar}{ma_0r} - \frac {\hbar}{2ma_0^2}$$

Hi, I still don't get it (I'm really sorry :(! ): when I replace ##a_0## by 2, the expression inside the brackets is still dependent on ##r##

Hi PeroK, thanks for replying. The solution would be ##a_0 =2## but I'm not sure I understand why I need this equation.

Thanks nqred, but I think I'm still missing something. Can this expression for ##a_0## be found purely mathematically or do I need to make assumptions/ rewrite other terms?

Hi, I think that this is my problem: I don't understand why the total energy/expression the brackets has no dependency on the variable ##r##.

Hi, thank you so. much for your reply, but I'm not sure I understand your point. Does this mean that ##a_0## has to be such that the expression in the brackets is equal to the sum of the potential and kinetic energies?

I am unable to complete the first part of the question. After I plug in the function for psi into the differential equation I am stuck: $$\frac {d \psi (r)}{dr} = -\frac 1 a_0 \psi (r), \frac d{dr} \biggl(r^2 \frac {d\psi (r)}{dr} \biggr) = -\frac 1 {a_0}\frac d {dr} \bigl[r^2 \psi(r) \bigr] =...