Determine the most probable radius for a 2s orbital (Hydrogen atom)

[science_1999]

Homework Statement

Determine the most probable radius for a 2s orbital (Hydrogen atom)

Homework Equations

Wavefunction for a 2s orbital:

1/(4√2pi*a^(3⁄2) ) (2-r⁄a) e^((-r)⁄(2a)) where, a=bohr radius

The Attempt at a Solution

First step:
find the probability density by squaring the wavefunction and multiplying but the spherical element of Volume

Second step:
Set the derivative of the probability density equal to zero to solve for where the slope of the plotted function is equal to zero

 

[nrqed]

Homework Statement

Determine the most probable radius for a 2s orbital (Hydrogen atom)

Homework Equations

Wavefunction for a 2s orbital:

1/(4√2pi*a^(3⁄2) ) (2-r⁄a) e^((-r)⁄(2a)) where, a=bohr radius

The Attempt at a Solution

First step:
find the probability density by squaring the wavefunction and multiplying but the spherical element of Volume

Second step:
Set the derivative of the probability density equal to zero to solve for where the slope of the plotted function is equal to zero

So you know all the steps, you simply have to go ahead!

Are you stuck on something?

 

[gijoe]

i am working on the same problem, and i am stuck on one particular part, when i take the derivative and set it equal to zero i end up with a cubic polynomial, like

(r/a)^3 - 6(r/a)^2 + 8r/a - 4 =0

not sure it this is correct, and if it is then do i need to use a graphing program to calculate the min and max points (roots) or is there an easier way?

thanks in advance for any help...

 

[Ed Aboud]

Stuck on this problem also!

I get to a stage where I get a massive cubic equation...I'm assuming I've gone wrong somewhere...

Any help? Or any online resources where the problem is worked through...?

 

[gijoe]

think i ended up solving it sometime last year.. think the equation factors to something like: (x-4)*(x-2)*(x-2)=0 or something like that.. roots are then 4 & 2.. can't remember exactly but hope this points you in the right direction... oh x=r/a) of course..