GR 8677927796770177 | # Login | Register

GR8677 #48
Problem
 GREPhysics.NET Official Solution Alternate Solutions

Atomic$\Rightarrow$}Transition

Transitions go from state one to another state that's completely unique relative to the first state. If both initial and final states have spherically symmetrical wave functions, then they have the same angular momentum. This is forbidden.

Alternate Solutions
 casseverhart132019-09-05 06:23:55 Many thanks conducive towards the great problem. click here\r\nReply to this comment jgramm2014-09-23 21:58:30 If they're both spherical, they're both in the s shell (think back to those silly pictures of orbitals in high school chem). If $\psi_1$ transitioned to $\psi_2$, both electrons (fermions) would occupy the same state, which is not allowed by the Pauli exclusion principle. Thus, D is the obviously the answer without having to worry about any of that selection rule nonsense that nobody ever remembers (or maybe it's just me).Reply to this comment
casseverhart13
2019-09-05 06:23:55
 danielsw986672019-10-21 06:53:45 D is the most obvious answer. altera
ernest21
2019-08-23 02:02:11
The information you share is very useful. It is closely related to my work and has helped me grow. Thank you! tapiriik
 danielsw986672019-10-21 07:00:03 It is indeed a piece of very useful information especially in physics. Cincinnati Property Management
joshuaprice153
2019-08-09 06:32:28
Thanks a lot for sharing us about this update. Hope you will not get tired on making posts as informative as this. computer repair Ocala
idoubt
2015-10-19 06:55:38
I think the answer has to do with the fact that photons are a spin 1 particle and it cannot have the m = 0 state. So to emit a photon the two wave functions must have an angular momentum difference $\\\\\\\\Delta l = \\\\\\\\pm 1$. This is not possible if both are S shell.
jgramm
2014-09-23 21:58:30
If they're both spherical, they're both in the s shell (think back to those silly pictures of orbitals in high school chem).

If $\psi_1$ transitioned to $\psi_2$, both electrons (fermions) would occupy the same state, which is not allowed by the Pauli exclusion principle.

Thus, D is the obviously the answer without having to worry about any of that selection rule nonsense that nobody ever remembers (or maybe it's just me).
 Baharmajorana2014-10-15 06:04:04 I agree with u, I can't remember too
 James6M2014-10-15 20:36:49 Um, what? There is only one electron in a hydrogen atom, so I don't know why you are talking about "both." And just because both wavefunctions are s-shells, that doesn't mean they are the same wavefunction. They can have different n. So, no, the Pauli exclusion principle has nothing to do with this question.
a_coiled_atom
2011-07-16 12:35:34
I thought of it in terms of the selection rules , . If the initial state has , which corresponds to spherical symmetry in the wavefunction, then according to these the final state must have , which is not spherically symmetric.
a_coiled_atom
2011-07-16 12:34:28
I thought of it in terms of the selection rules $l\to l\pm 1$, $m\to m\pm 1, m$. If the initial state has $l=0$, which corresponds to spherical symmetry in the wavefunction, then according to these the final state must have $l=1$, which is not spherically symmetric.
 pam d2011-09-24 11:31:55 This is how it's done.
thebigshow500
2008-10-13 22:12:48
Can anyone explain if other choices are something to do with wave function? (e.g. parity, orthogonality, etc.) Or are these choices just simply nonsense? :(
 kroner2009-10-16 12:06:41 (A): I'm not really sure what this is talking about. (B): Any two different eigenstates are mutually orthogonal. So in any transition this is required. (C): The eigenstates that are zero at the center are those with l>0. This is allowed (for example l could change from 3 to 2). (E): The selection rules require l to change.
 alemsalem2010-09-21 04:32:38 the states with even l have even parity(1) and those with odd l have odd parity(-1) since in transitions l changes by plus or minus 1 the parity must change in any transition, so (A) is out
lattes
2008-10-13 09:18:54
it might be useful to make an analogy with electromagnetism. It's known that a pulsating sphere does not emit radiation because of its spherical symmetry. An analogy can be made with wave-function. (D). For further reading check out Beiser's book on modern physics pag. 165.
curie
2008-10-08 05:55:58
Why does spherically-symmetric imply same angular momentum?
 rkraman2008-10-15 14:35:02 This has to do with the symmetry properties in mechanics. If the Lagrangian (and hence the Hamiltonian) describing a system is symmetrical with respect to any coordinate, then the canonical momentum corresponding to that coordinate is conserved. So if there is translation symmetry in $x$, the the canonical momentum which is the linear momentum $p$ is conserved. Similarly if the Hamiltonian is rotationally invariant, it is symmetrical with respect to $\theta$, then the corresponding angular momentum $\dot{\theta}$, which is nothing but the angular momentum $l$, is conserved. This is a specific instance of a fundamental theorem in Physics called the Noether's theorem.
 spacemanERAU2009-10-15 18:04:53 VERY VERY good explanation rkraman! kudos!
 flyboy6212010-11-14 20:52:31 Also, if the wavefunction is spherically symmetric, that means $l=0$.
evanb
2008-06-20 11:56:30
"completely unique relative..."

I'm not sure exactly what that means, but if it means that m must also change that's not quite right.

The selection rules are that NO TRANSITION OCCURS UNLESS:

$\Delta$m = {-1, 0, +1} and
$\Delta$l = {-1 , +1}.

$m$ has $m$ore options.

LaTeX syntax supported through dollar sign wrappers $, ex.,$\alpha^2_0$produces $\alpha^2_0$. type this... to get...$\int_0^\infty$$\int_0^\infty$$\partial$$\partial$$\Rightarrow$$\Rightarrow$$\ddot{x},\dot{x}$$\ddot{x},\dot{x}$$\sqrt{z}$$\sqrt{z}$$\langle my \rangle$$\langle my \rangle$$\left( abacadabra \right)_{me}$$\left( abacadabra \right)_{me}$$\vec{E}$$\vec{E}$$\frac{a}{b}\$ $\frac{a}{b}$