Calculating the probability of finding an electron

Click For Summary
The discussion revolves around calculating the probability of finding a 1s electron within a specific radius using the provided probability density function. Participants explore integrating the function over defined limits to find the probability within a sphere of radius a_o. One user successfully transforms the variable for easier integration and applies integration by parts, ultimately arriving at a numerical result. Concerns arise regarding the validity of an 11-digit probability result, as it should not exceed 1 due to normalization. The conversation emphasizes the importance of verifying solutions independently rather than relying solely on external sources.
baldywaldy
Messages
19
Reaction score
0

Homework Statement


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

probability = (4/a^3_o) r^2 e^{-2r/a_o} dr

1. work out the probability that an electron would be found in a sphere of radius a_o

Homework Equations



I know to find the probabilty you work out \int|\psi|^2 dr but because the probability is already given what do i do?

The probability is given in a region between r and r+dr so i guess I somehow work it out over 360 degress?


Thanks
 
Last edited:
Physics news on Phys.org
You have been provided with the probability density function - you need the probability between some limits... how would you normally do that? Think back to probability and statistics work you did in math class.
 
Simon Bridge said:
You have been provided with the probability density function - you need the probability between some limits... how would you normally do that? Think back to probability and statistics work you did in math class.

are the limits ao and 0 and then intergrate by parts?

Okay so if what i said is true:

I changed the variable to make it more straight forward (x=r/ao) so
(upper limit is 1 lower is 0)

probability = (4/a_o) x^2 e^{-2} dx
Then using intergration by parts I get

(4/a_o) \int x^2 e^{-2x} dx =
= 4/a_o [ (-(1/2)x^2 e^{-2x} - \int (-1/2)2x e^{-2x} dx) ]
= 4/a_o [(-(1/2)x^2 e^{-2x} - (1/2) x e^{-2x} - \int -(1/2) e^{-2x} dx)]
=4/a_o [(-(1/2)x^2 e^{-2x} - (1/2) x e^{-2x} - (1/4) e^{-2x} )]

so substituting in the upper and lower limits (1 and 0) i get

=(1/a_o) -(5e^{-2}/a_o)

If someone could check this for me or tell me where I have gone wrong. The problem i find is that the probability is a 11 digit number
 
Last edited:
I can help you check your own solutions.
Why is an 11 digit answer a problem?
 
Simon Bridge said:
I can help you check your own solutions.
Why is an 11 digit answer a problem?

Because the wavefunction is normalized it should be a maximum of 1 :). I've solved it now anyway, this website helped confirm my answer if anyone has a similar problem

Code:
http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hydrng.html#c1
 
Cool, well done.

Just some general notes:

You should be getting used to reasoning out your answers rather than relying on some outside authority. That's why I was being cautious about saying "yep - that's how you do it". How do you know I'm right? How do you know that website is right - maybe someone made a mistake?

Scientists can come across as very arrogant but one of the humilities in science is this distrust of argument by authority. A Nobel-Prize-winner can be challenged on the same grounds as anyone else.

0.12345678901 is an 11 digit number less than 1. See why I asked why an 11 digit answer was a problem?

You can make hyperlinks by putting the urls in tags. If you just paste the link i...phy-astr.gsu.edu/hbase/quantum/hydrng.html#c1

But you can also manually type them in for tidier links like this.
(Assuming you are using the quick reply box.)
 

Similar threads

  • · Replies 27 ·
Replies
27
Views
5K
Replies
6
Views
2K
Replies
7
Views
2K
  • · Replies 21 ·
Replies
21
Views
2K
Replies
5
Views
2K
  • · Replies 19 ·
Replies
19
Views
2K
Replies
14
Views
3K
  • · Replies 8 ·
Replies
8
Views
648
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 6 ·
Replies
6
Views
533