[poonilization]

so i'm a little unsure about this question http://i.gyazo.com/366ba4435170cda6ea7a121ea61cd1ec.png

for part i) , Ψ²= x⁴e^-x/a₀

then integrate using the hint, n=4 and a= 1/2ao i get 24a_o⁵ , substituting ao= 0.529Å i get 0.994 which is basically 1.
 So im not sure how do i normalize this wavefunction when the integral of Ψ² = 1

on part ii) i am kinda clueless on how to even attempt it. Ψ²= r²sin²θcos²Φe^-r/ao
How on earth would i integrate that, would i integrate everything individually so,

r² would be 1/3r³
then integrate sin²θ using double angle formulas
then integrate cos²Φ using double angle formulas
then finally integrate e^-r/ao to -ao/r e^-r/ao

and where/how would i use the hint dτ=r²drsinθdθdΦ

Any help would be much appreciated.

[Corribus]

If the integral of Ψ² is already equal to 1, then it's telling you the wavefunction is already normalized. That said, since you don't really have a unit for x, it is customary to leave a₀ as a constant and express the normalization factor in terms of a₀ (and, often, Z, the nuclear core charge, for hydrogenic wavefunctions). This way the normalization factor holds regardless of the length unit you choose to adopt. In other words, the fact that N is nearly equal to one when you substitute in for the Bohr radius (in Angstroms) is only coincidental. If you expressed it in, say, femtofurlongs or some such, this probably wouldn't happen.

For the second one, when you express your wavefunction, you should include your normalization factor that you are solving for (N). That's to start. Next, you need to realize that when you integrate in polar coordinates over all space, you don't just formulate the integral as dr dΦ dΘ, because this isn't an infinitesimal volume element in polar coordinates. There is an extra value of r² and sin(θ) that have to be added to the integrand. This is important!!! To see why, you need to practice converting between Cartesian and polar coordinate systems. This is what the hint is referring to. Finally, yes you can separate the variables in this case, which makes solving the integral much easier.

[poonilization]

So for part i) Ψ²= A²x⁴e^-x/a0

then integrate where n=4 and a=1/2ao to give me
 A²24ao⁵=1

rearranging for A

A = square root of 1/24ao⁵

So this is the normalization factor.


for part ii) integral of Ψ²=A² r²sin²θcos²Φe^-r/ao dτ = 1

subbing in dτ=r²drsinθdθdΦ

integral of Ψ²=A² r²sin²θcos²Φe^-r/ao r²drsinθdθdΦ = 1
simplifying this: A² r⁴e^-r/aodr sin³θdθ cos²ΦdΦ  = 1

=A² r⁵/5 *-ao/r e^-r/ao * 1/3cos³θ - cosθ * 1/2 + 1/2sinΦcosΦ =1

simplifying = A² -r⁴/5 * ao e^-r/ao * 1/3cos³θ - cosθ * 1/2 + 1/2sinΦcosΦ =1

where do i go from here, that is assuming if i have done everything up till now correctly. Would i just substitute in the limits in the equation above?

So A² ([equation] with limits r = infinity , θ = pi and Φ = 2pi)  - ([equation] with limits r = 0 , θ = 0 and Φ = 0) = 1

[Corribus]

Part 1 looks like what I got.
Part 2, not really sure what you're doing with the math. You set up the integral right (line that start with "simplifying this".  Now solve each part separately, and multiply the results together, solve for A. You can look up the trigonometric integrals at online tables, such as here:

http://integral-table.com/downloads/single-page-integral-table.pdf

Note that the integral for r is exactly what you solved in part 1, so you really only have two fairly simple integrals to solve (simple if you look them up in integral tables, and use the integration limits).

[poonilization]

ok thanks for the reply

so continuing from simplifying this: A² r⁴e^-r/aodr sin³θdθ cos²ΦdΦ  = 1

integrating that i get: A² 24ao⁵ (1/3cos³θ - cosθ) (1/2Φ+1/4sin(2Φ)) =1

i'm not sure if i have integrated sin³θ correctly, when i check it on google i get (1/3cos³θ - cosθ) from multiple websites/forums/yahoo answers etc however the table you linked states it would be -3/4cosθ + 1/12cos3θ. Im guessing they are the same but just represented differently.

substituting the limits A² [24ao⁵ (1/3cos³pi - cospi) (1/2(2pi)+1/4sin(2(2pi))] - [24ao⁵ (1/3cos³0 - cos0) (1/2(0)+1/4sin(2(0))] =1

cospi = -1
sin2pi = 0
cos0 = 1
sin0 = 0

A² [24ao⁵(-1/3 - (-1)) (1/2(2pi))] - [24ao⁵ (1/3 - 1)] =1

A² [24ao⁵(2/3) (pi)] - [24ao⁵ (-2/3)] =1

A² [16ao⁵pi] + [16ao⁵] = 1

rearranging for A

A=square root of (1 - 16ao⁵)/16ao⁵pi

[Corribus]

Ok, the integral tables seem to be equivalent, so no worries there. But I think you're still making a mistake somewhere in the math. The way I did it was to solve each integral completely independently, because the variables are completely separable, and then multiply everything together at the end.