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Probability to find electron in ground state

  1. Jan 28, 2014 #1
    1. The problem statement, all variables and given/known data
    An electron in hydrogen atom can be described with normalized wavefunction ##\psi (r,\vartheta ,\varphi )=(\frac{1}{64\pi r_B^3})^{\frac{1}{2}}exp(-\frac{r}{4r_B})##.

    Calculate the probability that the electron is in ground state of hydrogen atom. How much is the energy of electron in state ##\psi ##?
    Ground state is ##\psi (r,\vartheta ,\varphi )=\frac{1}{\sqrt{4\pi }}\frac{2}{r_B^{3/2}}exp(-\frac{r}{r_B})##


    2. Relevant equations



    3. The attempt at a solution

    I have no idea on how to even start with the fist part - calculating the probability...

    But the second part should be relatively easy to do: ##\left \langle \psi\left | \hat{H} \right | \psi \right \rangle=\left \langle \psi\left | \frac{\hbar^2}{2m}\hat{p}^2+\frac{\hat{l}^2}{2m\left \langle r \right \rangle^2}-\frac{e^2}{4\pi \varepsilon _0}\frac{1}{\hat{r}} \right | \psi \right \rangle##

    But still... what is the idea behind the first question? o_O
     
  2. jcsd
  3. Jan 28, 2014 #2

    TSny

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    Let ##\psi## be your given normalized wavefunction and let ##\phi_1, \phi_2, \phi_3,....## be the normalized wavefunctions for the energy eigenstates of the atom. So, ##\phi_1## is the ground state, ##\phi_2## is the first excited state, etc.

    The set of eigenfunctions ##\{\phi_1, \phi_2, \phi_3,...\}## form a complete, orthonormal set of functions. So, you could imagine expanding ##\psi## in terms of the eigenfunctions:

    ##\psi = a_1\phi_1+a_2\phi_2+a_3\phi_3...##

    How is the coefficient ##a_1## related to the probability that you are asked to find, and how can you calculate the value of ##a_1##?
     
  4. Jan 28, 2014 #3
    ##\left | a_1 \right |^2## is by definition the probability that the electron will be in state ##\phi _1##.

    The sum of those coefficients is therefore ##\sum_{i=1}^{n}\left | a_i \right |=1##.

    Now how to calculate them... hmm

    If I am not wrong ##a_n=\left \langle \psi _n \right | \psi \rangle## or is it not?
     
  5. Jan 28, 2014 #4
    ##a_1=\left \langle \psi _1 \right | \psi \rangle=\int_{0}^{2\pi }d\varphi \int_{0}^{\pi }d\vartheta sin\vartheta \int_{0}^{\infty}(\frac{1}{64\pi r_B^3})^{\frac{1}{2}}exp(-\frac{r}{4r_B})\frac{1}{\sqrt{4\pi }}\frac{2}{r_B^{3/2}}exp(-\frac{r}{r_B})r^2dr##

    ##a_1=\frac{1}{2r_B^3}\int_{0}^{\infty}r^2exp(-\frac{5r}{4r_B})dr##

    which gives me ##a_1=\frac{16}{25}##

    Therefore the probability should be 0.4096.

    I hope.
     
  6. Jan 28, 2014 #5

    TSny

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    Looks good.

    I get a different value.
     
  7. Jan 28, 2014 #6
    Of course you do, ##5^2## was never ##20##.

    ##a_1=64/125##.

    Thank you TSny!
     
  8. Jan 28, 2014 #7

    TSny

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    :smile:
     
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