[JohnTan]

Hi everyone,
I am a little confused about one of the questions in my book that says 'the square of the wavefunction gives the probability of finding a particle at a given point.' - this i understand, however, it goes on to ask 'what is the probability of finding an electron in a 1s orbital within a volume of 1pm^3 centered (a) at the nucleus and (b) 50pm away from the nucleus?' I assume the answer to a is 0, however I'm not sure about how to go about b.
Thanks for any help

[nigel433]

If you are quoting accurately, the book is wrong.

Not the "probabiity" but the "probability DENSITY".

The distinction is nothing to do with quantum mechanics specifically
but applies in probability theory whenever you have a continuous
variable. You can't really talk about the probability AT a point but
you can talk about the probability in a defined space CONTAINING
the point;and such talk needs some sort of measure of the local
intensity of the phenomenon - which is this psi squared.

Congratulations if you "understand" your book.
The talk of "probability of finding an electron" is just hand-waving.
Nobody. and I mean nobody - ever - in history has had the slightest
idea what psi, is or means. For some reason, something on the nano
scale obeys mathematical rules which are similar to the ones which
describe standing waves on vibrating metal plates (Chladni patterns,
discovered 1840).

"If you think you understand quantum mechanics, you are wrong.
Nobody does"; Richard Feynman, Nobel Laureate in Physics.

Anyway:

The orbitals are not like planetary orbits in which the planet travels
on imagined lines and never comes near the sun. Orbitals are imagined
volumes where the electron may be, and the first s-orbital is a sphere
centered on the nucleus (and including the nucleus; yes, experiments
show that electrons DO  "spend time" IN the nucleus - and yet the world
does not blow up.) So the answer to Part (a) is not zero but some number
which you have to calculate.

As I say, Part (b) needs a calculation of the probability density at
the given radius. Since an s-orbital is spherical, the rule for calculations
is symmetrical about the nucleus and very simple (although I do not have it
to hand)

[nigel433]

I have looked again at your post and I notice that Part (b)
says "50 pm away". The density is going to be close to zero
out there,I think.

I guess the question may be asking you to use your common
sense and get a feel for where the electron is localised and where
it hardly ever is. Still, you had better calculate it and not trust to
my intuition.