The following is a translation from the :  The ionisation energy or binding energy B is the sum of potential energy V and kinetic energy T of the electron. $B = V+T$. Since the following always applies in a stable orbit: $2 T = -V$, the ionisation energy is always equal in magnitude to the kinetic energy or half the potential energy: $B = T+V = -T = V/2$.  Apart from the fact that the 13.6 eV for the ionisation energy of hydrogen was obtained from the assumption of an orbiting electron, I do not understand why $2 T = -V$?
120829
I came across this relation while reading on the Bohr atomic model. Are there any other forces for which these relations hold good?
65433
Why that exact number of electrons in one coulomb? who decided it? there is nothing wrong with the number, it just seems slightly messy. Why didn't the scientific community just settle on an easier number, such as $1\times10^{-19}$ for example?
22936
So, I was reading Atom: Journey Across The Subatomic Cosmos by Isaac Asimov in order to better understand quantum mechanics when I came across this sentence:      The electron couldn't spiral into the proton because it couldn't take up an orbit with a length less than a single wave.    I was wondering why that is? Why can't an electron not have non-whole number wavelengths in atomic energy levels?
1968
One of the basic experiments everybody learns about when it comes to physics is the spectroscopy of hydrogen gas. The typical set-up has a charge discharge lamp containing the hydrogen. The resulting spectra are then discussed with the atomic properties of the hydrogen atom.  The gas however is not mono-atomic but it is present as $\textrm{H}_{2}$. The 1s orbitals creating the $\sigma$-bond and the electrons are occupying the bound state. This is at a lower energy when compared to the 1s states of the atomic hydrogen. The higher energy orbitals will also influence each other as they are overlapping. So my question is, why does the atomic hydrogen description works for this experiment?  Possible answers I could think of are:   The bounding effects are negligible The $\textrm H_{2}$ spectra are only visible shortly after the lamp is switched on. The constant electron bombardment is slowly dissociating all $\textrm H_{2}$.   For the later I am very unsure if a charge discharge lamp is able to produce conditions to archive that. If it comes out that this is the answer, does anyone know where to find more details on the spectroscopic properties of $\textrm H_2$?
68259
Since energy can always be shifted by a constant value without changing anything, why do books on quantum mechanics bother carrying the term $\hbar\omega/2$ around?  To be precise, why do we write $H = \hbar\omega(n + \frac{1}{2})$ instead of simply $H = \hbar\omega n$.  Is there any motivation for not immediately dropping the term?
70429
In the derivation of drift velocity I have seen two variations and want to know which one's correct.   $s=ut+\frac{at^2}{2}$ Assume that the drift velocity of any electron in any conductor is : $$\vec{v_d}=\frac{\vec{l}}{t}$$ Due to the electric field the acceleration of electrons in any conductor is: $$a=\frac{-e\vec{E}}{m}$$ Now the distance travelled by an electron after a long time (initial thermal velocity = 0) $$l=\frac{at^2}{2}\implies \frac{-e\vec{E}t^2}{2m}$$ the time between the collisions is $\tau$ $\therefore$ $$l=\frac{-e\vec{E}\tau^2}{2m}$$ thus the velocity is $$\vec{v_d}=\frac{-e\vec{E}\tau}{2m}$$ In another proof I saw the author using $v=u+at \implies v_d=\frac{-e\vec{E}\tau}{m}$   My question is which of the the two equations of motion can be used in the proof? Can they be used at all.
66560
I have been reading through this  of the ground state energy of hydrogen  and others like it. In this one it says it is using the uncertainty principal but then proceeded to use the following: $$pr=\hbar$$ But why is it not using: $$pr=\frac{\hbar}{2}$$ which is more in line with the uncertainty principle and what comes earlier in the derivation.
74471
Why do we use the ?  Why did it come to be the electron volt and not, say, just a prefix of the joule, like the nanojoule?  Does the electron volt represent anything particular as far as the mathematics goes? I am guessing that it does, and if so, what is it that the electron volt exactly represents in terms of the mass of a particle, as I have seen it used for both the energy of a photon and the mass of subatomic particles?
97469
Say we have 2 charges set up like above. Why is $V_3=0$ ?   I can understand mathematically that the scalars fields cancel out, but I don't understand it physically. If I place a positive charge in between the 2 charges, it has electric potential energy because it is feels a force towards the negative charge and repelled by the positive and gains kinetic energy. However because the Electric Potential is 0, thus the electrical potential energy is 0.  How can $V_3 =0V$ if a charge placed in between the two charges will gain kinetic energy?
72052
I understand that the average energy of each degree of freedom in a thermodynamic system is $\frac12kT$. And so, for an ideal monatomic gas, there are three degrees of freedom associated with the translational components of each atom, which gives $E$ = $\frac32kT$ for each atom in this system.  For an ideal diatomic gas, there are the three degrees of freedom from the translational components plus two more degrees of freedom associated with the rotational components of each molecule. So we end up with $E$ = $\frac52kT$ for each molecule.  My question here is why did we ignore the rotational degree(s) of freedom in case of a monatomic gas ? Like why isn't there another degree of freedom for the rotational component of a single atom ? Also, if that is neglected for some reason, then when does it become significant and can no longer be ignored ? For example, shouldn't it be included in calculations under extreme conditions like temperature of millions of degrees ?
11990
Followup to "".  It is argued that even a tiny residual charge would result in huge amounts of electricity in bulk matter, everything would be different, etc. I do not find that a convincing answer: suppose $n$ protons plus $n+1$ electrons are neutral. Why wouldn't we also expect there to be $n$ protons to every $n+1$ electrons? That is, there is no bulk matter problem if every $n$'th atom is a negative ion (for hydrogen).  Now, what empirical lower bound can we give for $n$ in that kind of scenario?
66045
My original thought was:  $$ L = L_0 \left( 1 - e^{-t/\tau} \right) $$  where $L$ = luminosity and $L_0$ = initial luminosity.   But when I tested this I found that the time constant $\tau$ was higher for smaller values of Voltage, so I have no idea and all I can find is the luminosity of planets etc. Any help would be good.
59403
Is there a good physical picture of why the energy levels in a hydrogen atom are independent of the angular momentum quantum number $\ell$ and $m$?
50221
Suppose we have a particle of mass $m$ that is in an eigenstate $|\psi\rangle$ of the Hamiltonian $\hat{H}=\hat{T}+\hat{V}$, where $\hat{T}$ is the kinetic energy operator and $\hat{V}=V(\hat{r})$ is the potential energy operator. If the potential has lower bound $V_0$, then is it necessary for the energy eigenvalue $E$ of $|\psi\rangle$ to be greater than $V_0$? Classically this is true, since we regard negative kinetic energy as physically unrealizable/meaningless. However, I don't know if I can necessarily say the same in the quantum case. For example, what I'm tempted to do is write  $$\langle\psi | \hat{T}|\psi \rangle = \langle\psi | E-\hat{V}|\psi\rangle$$  and say "If $E&lt;V_0$, then the RHS is necessarily negative, implying the LHS is as well, which we will regard as physically meaningless. If $E=V_0$, then $|\psi\rangle$ is trivial.", but I'm not sure if that's right.   Confused by that, I then wanted to show that, if $E&lt; V_0$, then a non-trivial $|\psi\rangle$ is non-normalizable. However, I'm not entirely sure how to do this.
90760
The nuclear binding energy, is the energy that is needed to seperate the nucleons in a nucleus. And binding energy is also defined as the energy given out when a nucleus forms from nucleons.   Also the larger the nucleus is, the more energy is required to break it apart, so why doesn't that mean that larger nuclei are more stable? I mean Uranium has a lower binding energy per nucleon than Iron, but there are many many more nucleons in Uranium that Iron so the total binding energy is going to be much greater.   Basically I don't understand why whether an element gives out energy by fusion or fission (why the lighter element provide energy by fusion not fission and vice versa for heavy elements) depends on binding energy per nucleon and not "total" binding energy
88287
The solutions of the Schrödinger equation for hydrogen are the "electronic orbitals", shown in this picture:  ()  They have the following degeneracy structure:  ()  It is often said that atoms in other elements simply have more electrons "filling up" the orbitals in increasing order of energy, as required by the Pauli exclusion principle.  This is somewhat confirmed via xray spectroscopy:  ()  But why would non-hydrogen atoms have the same orbital degeneracy structure as hydrogen, if the actual Hamiltonian is different from one atom to another?
68738
First and foremost I am sorry for this; in order to make the questions clear I must first upload the lecture notes from my institution $^\zeta$ for the derivation of Hydrostatic Equilibrium:          $^\zeta$ Lecture notes courtesy of Imperial College London, Astrophysics dept, 2017-2018 edition.    Firstly, in equation $(3.1)$ why is there no negative sign for $F_g$? Last time I checked, Newton's law of gravitation told us that   $$F=-\frac{GMm}{r^2}$$  Nextly, I have a conceptual problem with equation $(3.2)$; when they use the term 'pressure' I am going to assume that they are referring to the radiation pressure that is released due to nuclear fusion reactions taking place in the sun. In that formula it must be the case that $P(r) \gt P(r+\delta r)$ if this is the case then $(3.2)$ should read $$F_p=-\left(\frac{dP}{dr}\right)\delta r\,\delta A $$   Correct me if I'm wrong but Hydrogen nuclei are fused together to generate the more stable Helium and hence lots of radiation pressure acting outwards from the centre of the star. If this is correct then $P(r+\delta r)=0$, since the outside face of that cylindrical volume element in Figure $3.1$ will not 'feel' any of this radiation pressure because this pressure only acts on the inner face. What am I not understanding here?  Moving on to equation $(3.3)$, let's suppose the sign of the gravitational force in $(3.1)$ really is neglected. Does this mean that the negative sign in $(3.3)$ is due to $P(r+\delta r) \lt P(r)$ and hence $$\frac{GM(r)\rho(r)}{r^2}=-\left(\frac{dP}{dr}\right)\,?$$ I know this is the same equation as in $(3.3)$ but I have written it with the negative sign next to the pressure differential as I would like to know where this minus sign originates from. Does anyone know?  My final concern is the sign in $(3.4)$ as last time I checked $$g=-\frac{GM}{r^2}$$ and not $$g=\frac{GM}{r^2}$$  Hence, equation $(3.4)$ should read  $$\left(\frac{dP}{dr}\right)=\rho(r)\,g\,?$$     I forgot to mention that I have done some research and  is the closest derivation I could find, which, needless to say is no less helpful than the notes I have.