10.5: Problems
- Page ID
- 106866
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.
- Calculate \(A\)
- calculate \(\left \langle r\right \rangle\), the average value of the distance of the electron from the nucleus.
- 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.
- What is the probability that the electron is found at a distance from the nucleus equal to \(a_0/2\)?
- What is the probability that the electron is found at a distance from the nucleus less than \(a_0/2\)?
- 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!
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.
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