# Calculating the probability of finding an electron

1. Nov 19, 2011

### baldywaldy

1. The problem statement, all variables and given/known data
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 $$a_o$$

2. Relevant 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: Nov 19, 2011
2. Nov 19, 2011

### Simon Bridge

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.

3. Nov 19, 2011

### baldywaldy

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: Nov 19, 2011
4. Nov 21, 2011

### Simon Bridge

Why is an 11 digit answer a problem?

5. Nov 23, 2011

### baldywaldy

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

Code (Text):
http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hydrng.html#c1

6. Nov 23, 2011

### Simon Bridge

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.)