schrodinger stuff
need some clarity over psi(r) , psi^2(r) and 4pir^2 psi^2(r) i.e P(r) graphs.
some1 pls specify how to identify which graphs corresponds to which orbital
some1 pls specify how to identify which graphs corresponds to which orbital
I@Dexter
INote for OP
+solved @user1 @user2...
W+1
Wmujhe bhi batao
Afrom my observations i have seen psi(r) graph is like a wave (uppar neeche jata rehta hai), psi^2(r) since its square will always be above the x axis P(r) always starts from origin because it has r term and we can use radial nodes = n-l-1 for all 3
Acan anyone else cnf if this is correct
Adk about any significance of psi(r) but psi^2r is probability density and 4pir^2dr times this will give just the probability (radial) since were doing density*volume
Athats why for s psi^2 starts from a maxima cuz at centre the prob of finding e- is finite but its zero for others
IIt’s there in my theory videos. Check it out.
IAlso, all graphs are there in Atkins. Some are in jd lee too.
Also present in zumdahl.
Also present in zumdahl.
V$\psi(r)$ is the equation of the wave that the electron corresponds to. It's the Radial Wave Function
$|\psi(r)|^2$ is a part of the actual term that represents probability that's actually $4\pi|\psi(r)|^2dr$. This is called the Radial Distribution Function
$|\psi(r)|^2$ is a part of the actual term that represents probability that's actually $4\pi|\psi(r)|^2dr$. This is called the Radial Distribution Function
TEnamine
VThe difference in their graphs is just that 'cause the term has been squared, any values of r for which $\psi(r)$ is negative becomes positive and comes above the x axis
TEnamine
VThere is this one small thing that you'll kinda need to remember (altho a good explanation does exist)
The graph of $\psi(r)$ vs $(r)$ for an s orbital starts from a non zero value of $\psi(r)$ while it starts from zero for all the other ones i.e. the ones that have an $l>0$
The graph of $\psi(r)$ vs $(r)$ for an s orbital starts from a non zero value of $\psi(r)$ while it starts from zero for all the other ones i.e. the ones that have an $l>0$
TEnamine
VThe reason of it lies in the fact that the orbital angular momentum of any electron lying in the s orbital is zero by the following : $l = \hbar\cdot \sqrt(l(l+2)$
TEnamine
VSmall Correction: it should be $p\pi|\psi(r)|^2r^2dr$
VSmall Correction: it should be $4\pi|\psi(r)|^2r^2dr$
TEnamine
V$maja\aa\raha\tha\ \LaTeX\ me\ likhne\ me$
TEnamine
Iलोल kyu bacche ki zindagi barbaad kar rahe ho :p
IGreat point. Pehle s orbital toh samajhne dijiye :)
Cok so from discussion i ve come up with the conclusion
psi(r) represents the wave / amplitude
psi^2(r) represent the probability density
integration of 4pir^2 psi^2(r) dr gives radial probability function
psi(r) represents the wave / amplitude
psi^2(r) represent the probability density
integration of 4pir^2 psi^2(r) dr gives radial probability function
Cno of points where psi(r) vs r cuts x axis = no of radial nodes
Cno of points where psi^2(r) vs r touches x axis = no of radial nodes
Cno of points where P(r) vs touches x axis (except origin) = no of radial nodes
Cno of peaks in P(r) vs r = total no of nodes + 1 = quantum number
ILast maxima of probability gives you radius.
IThis is same as the formula every one mugs :) total nodes is n-1
INicely done
CW
Cthanx
Vhahaha
VI mean s orbital samajhne ke liye to mujhe bhi ek semester hai abhi 
VI know the reasoning about angular momentum that I gave is kinda circular but that's the best I can offer without delving into the math of it
IYea lots of math.
IAtkins is nice for this
Vhmm
VI'm going to study quantum chem soon enough anyways
Vso doesn't matter I guess lol
C+solved @VarunArora @vj25
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