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Chemistry LibreTexts

10.5: Problems

  • Page ID
    106866
  • Problem \(\PageIndex{1}\)

    The wave function describing the state of an electron in the 1s orbital of the hydrogen atom is:

    \[\psi_{1s}=Ae^{-r/a_0}, \nonumber\]

    where \(a_0\) is Bohr’s radius (units of distance), and \(A\) is a normalization constant.

    1. Calculate \(A\)
    2. calculate \(\left \langle r\right \rangle\), the average value of the distance of the electron from the nucleus.
    3. The radius of the hydrogen atom is taken as the most probable value of \(r\) for the 1s orbital. Calculate the radius of the hydrogen atom.
    4. What is the probability that the electron is found at a distance from the nucleus equal to \(a_0/2\)?
    5. What is the probability that the electron is found at a distance from the nucleus less than \(a_0/2\)?
    6. We know that the probability that the electron is found at a distance from the nucleus \(0 < r < \infty\) is 1. Using this fact and the result of the previous question, calculate the probability that the electron is found at a distance from the nucleus greater than \(a_0/2\).

    Hint:\(\int x^2 e^{ax}dx=e^{ax}\frac{\left ( 2-2ax+a^2x^2 \right )}{a^3}\)

    Note: Be sure you show all the steps!

    Problem \(\PageIndex{2}\)

    The wave function describing the state of an electron in the 2s orbital of the hydrogen atom is:

    \[\psi_{2s}=Ae^{-r/2a_0}\left(2-\frac{r}{a_0}\right) \nonumber\]

    where \(a_0\) is Bohr’s radius (units of distance), and A is a normalization constant.

    • Calculate \(A\)
    • Calculate \(\left \langle r\right \rangle\), the average value of the distance of the electron from the nucleus.

    Problem \(\PageIndex{3}\)

    Calculate the normalization constant of each of the following orbitals:

    \[\psi_{2p+1}=A_1 r e^{-r/2a_0}\sin \theta e^{i\phi} \nonumber\]

    \[\psi_{2p-1}=A_2 r e^{-r/2a_0}\sin \theta e^{-i\phi} \nonumber\]


    1The integral in \(r\) was solved using the formula sheet

    2If you find this strange think about a situation where 20 18-year olds gather in a room with 4 60-year olds. The average age in the room is 25, but the most probable age is 18

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