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Homework Help: Calculating the probability of finding an electron

  1. Nov 19, 2011 #1
    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 = [tex] (4/a^3_o) r^2 e^{-2r/a_o} dr [/tex]

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

    2. Relevant equations

    I know to find the probabilty you work out [tex]\int|\psi|^2 dr[/tex] 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?

    Last edited: Nov 19, 2011
  2. jcsd
  3. Nov 19, 2011 #2

    Simon Bridge

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    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.
  4. Nov 19, 2011 #3
    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 = [tex] (4/a_o) x^2 e^{-2} dx [/tex]
    Then using intergration by parts I get

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

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

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

    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
  5. Nov 21, 2011 #4

    Simon Bridge

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    I can help you check your own solutions.
    Why is an 11 digit answer a problem?
  6. Nov 23, 2011 #5
    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):
  7. Nov 23, 2011 #6

    Simon Bridge

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    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.)
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