Integrating by Parts: Solving for Probability in Sphere of Radius a0

[unknownuser9]

Homework Statement

Given that the probability of finding a 1s electron in a region between r and r + dr is:

P = $$\frac{4}{a_{0}^{3}}$$r$$^{2}$$e^{[tex]-2r/a₀[/tex]}dr


work out the probability that an electron would be found within a sphere of radius:

i) a₀

Homework Equations

The Attempt at a Solution

It hints that substituting x = r/a0 makes it easier to integrate by parts.

So, substituting and cancelling:

P = 4x² e^-2x dx

Integrating by parts:

P = (1/2)x²e^-2x + xe^-2x - (1/4)e^-2x

And then use 0 and a0 as my limits?

Thanks.

 

[Mark44]

Cleaned up your LaTeX.
Tips:
Don't put [ sup] and [ sub] tags inside LaTeX tags.
Use one set of [ tex] and [ /tex] tags for an entire expression.

Homework Statement

Given that the probability of finding a 1s electron in a region between r and r + dr is:

P = $$\frac{4}{a_{0}^{3}}r^2e^{-2r/a_0}dr$$


work out the probability that an electron would be found within a sphere of radius:

i) a₀

Homework Equations

The Attempt at a Solution

It hints that substituting x = r/a0 makes it easier to integrate by parts.

So, substituting and cancelling:

P = 4x² e^-2x dx

Integrating by parts:

P = (1/2)x²e^-2x + xe^-2x - (1/4)e^-2x

And then use 0 and a0 as my limits?

Thanks.

 

[Mark44]

Now that I can read the probability expression, I can provide some help. To find the probability of finding a 1s electron in a sphere of radius a₀, you're going to need to integrate that probability expression from r = 0 to r = a₀.

When you make the substitution x = r/a₀, dx = dr/a₀. I don't see dx in your integrand (or even an integral at all). You showed the results of integration by parts, but not the work, so unless I duplicate your work, I can't say that your work is correct.

 

[unknownuser9]

Sorry, i really can't use Latex. I did it in word and this is the best i can do:

P= ∫▒4/(a_0^3 ) r^2 e^((-2r)/a_0   ) dr
P= ∫▒4/(a_0^3 ) 〖a_0〗^2 r^2 e^(-2x ) a_0 dx
P= ∫▒〖4x〗^2  e^(-2x) dx
P=4∫▒x^2  e^(-2x) dx
P= -1/2 x^2 e^(-2x)+∫▒〖xe^(-2x) 〗 dx
P= -1/2 x^2 e^(-2x)+xe^(-2x) ∫▒1/2  e^(-2x) dx
P= -1/2 x^2 e^(-2x)+xe^(-2x)-1/4  e^(-2x)

So now if i use r = 0 and r = a_0 as my limits i get:

P= |-1/2 x^2 e^(-2x)+xe^(-2x)-1/4  e^(-2x) |  a_0¦0
P=(-1/2 a_0^2 e^(-2a_0 )+a_0 e^(-2a_0 )-1/4 e^(-2a_0 ) )- (0+0-1/4 e^0)
P=-1/2 a_0^2 e^(-2a_0 )+a_0 e^(-2a_0 )-1/4 e^(-2a_0 )+1/4

Am i close? Thanks.

 

[unknownuser9]

Ive just realized, I took 4 out as a constant at the beginning. Would i have to mulitply through by 4 now?

 

[Mark44]

Sorry, i really can't use Latex. I did it in word and this is the best i can do:

It's not that hard to use LaTeX, and it's much more readable that what you have below. For me the limits of integration show up as a rectangular pattern of dots, like this: ▒.
Bring the constants out of the integral.
$$\frac{4}{a_0^3}\int_0^{a_0}r^2e^{-2r/a_0}dr$$

To see what I did in LaTeX, click the expression above.

After you make the substitution, use integration by parts twice. After you integrate, undo your substitution. Check your final result by differentiating - you should get back to the original integrand.

P= ∫▒4/(a_0^3 ) r^2 e^((-2r)/a_0   ) dr
P= ∫▒4/(a_0^3 ) 〖a_0〗^2 r^2 e^(-2x ) a_0 dx
P= ∫▒〖4x〗^2  e^(-2x) dx
P=4∫▒x^2  e^(-2x) dx
P= -1/2 x^2 e^(-2x)+∫▒〖xe^(-2x) 〗 dx
P= -1/2 x^2 e^(-2x)+xe^(-2x) ∫▒1/2  e^(-2x) dx
P= -1/2 x^2 e^(-2x)+xe^(-2x)-1/4  e^(-2x)

So now if i use r = 0 and r = a_0 as my limits i get:

P= |-1/2 x^2 e^(-2x)+xe^(-2x)-1/4  e^(-2x) |  a_0¦0
P=(-1/2 a_0^2 e^(-2a_0 )+a_0 e^(-2a_0 )-1/4 e^(-2a_0 ) )- (0+0-1/4 e^0)
P=-1/2 a_0^2 e^(-2a_0 )+a_0 e^(-2a_0 )-1/4 e^(-2a_0 )+1/4

Am i close? Thanks.

 

[unknownuser9]

Yer I know but its never worked on my computer. Even if I copy and paste exactly what you wrote, It doesn't work.

OK thanks, I will try that.

 

[unknownuser9]

I have hosted an image of my calculations in word (LaTex still isn't working for me):

http://img221.imageshack.us/img221/8241/probof1selectron.jpg

Are my workings right?

When I substitute in the limts, I get negative numbers which I am sure is not right.

Thanks again.

 

[Mark44]

I'm pretty sure you have a sign error.
With the substitution x = r/a (I didn't bother with the subscript on a), I have this integral
$$4\int x^2 e^{-2x}dx$$
which evaluates to -2x²e^-2x - 2xe^-2x - e^-2x. I have a minus before the third term, and you have a plus.

Instead of undoing the substitution, you can change the limits of integration. The original limits were values of r, and were 0 and a. Since x = r/a, the new limits are 0 and 1. If you evaluate -2x²e^-2x - 2xe^-2x - e^-2x at 1 and 0, and subtract the two values, you get 1 - 5e^-2, which is about .323324, a seemingly reasonable value for a probability.

 

[unknownuser9]

Yeah, i realized i hadnt evaluated the final integral. When i did i got that result and substituting the limits gave me that answer. Thanks for your help!