- #1
cr7einstein
- 87
- 2
Hi everyone;
A very stupid confusion here. When we want to talk about the most probable radius to find the electron in $1s$ orbital, why do we talk about the radial density and not the probability itself? For instance, the probability of finding the the electron at a radial distance $r$ is($\rho$ being the square modulus of the radial part of wavefunction):
$$dP=\rho(r)dV=\rho(r)4\pi r^2dr=\rho '(r)dr$$, since the electron can be anywhere in the shell at radius $r$. Now, the probability of finding electron is maximum when $$dP/dr=0$$right? Isn't that the condition for maxima? But that gives $$r=0$$, i.e. the nucleus. Now I know that this is wrong, I'm supposed to get the Bohr radius.
However, and we all know this, extremising $$\rho'(r)$$ i.e. the radial density gives the correct answer.
What I don't understand is, why am I wrong the first way? After all, maximising(okay, extremising) anything is to put the first derivative to zero, right? Then what am I missing here?I can 'see' that this is wrong because $$dP/dr=0$$ means my radial density is zero, which is a little absurd, but that is not a good argument at all.
In a nutshell, why do we extremise the RADIAL DENSITY, and NOT PROBABILITY while finding the maximum probability? If we were to tell the most probable radius where this DENSITY is maximum, then I'd be fine with the second way. But if we were to find the greatest probability itself, isn't the second approach counter intuitive?
I know I am missing something very subtle; please point it out to me. Thanks in advance!
PS: I read up similar questions here, bu I feel my question hasn't exactly been answered in those.
A very stupid confusion here. When we want to talk about the most probable radius to find the electron in $1s$ orbital, why do we talk about the radial density and not the probability itself? For instance, the probability of finding the the electron at a radial distance $r$ is($\rho$ being the square modulus of the radial part of wavefunction):
$$dP=\rho(r)dV=\rho(r)4\pi r^2dr=\rho '(r)dr$$, since the electron can be anywhere in the shell at radius $r$. Now, the probability of finding electron is maximum when $$dP/dr=0$$right? Isn't that the condition for maxima? But that gives $$r=0$$, i.e. the nucleus. Now I know that this is wrong, I'm supposed to get the Bohr radius.
However, and we all know this, extremising $$\rho'(r)$$ i.e. the radial density gives the correct answer.
What I don't understand is, why am I wrong the first way? After all, maximising(okay, extremising) anything is to put the first derivative to zero, right? Then what am I missing here?I can 'see' that this is wrong because $$dP/dr=0$$ means my radial density is zero, which is a little absurd, but that is not a good argument at all.
In a nutshell, why do we extremise the RADIAL DENSITY, and NOT PROBABILITY while finding the maximum probability? If we were to tell the most probable radius where this DENSITY is maximum, then I'd be fine with the second way. But if we were to find the greatest probability itself, isn't the second approach counter intuitive?
I know I am missing something very subtle; please point it out to me. Thanks in advance!
PS: I read up similar questions here, bu I feel my question hasn't exactly been answered in those.