1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Expression for Bohr Radius using Quantum Mechanics; In Commutator Relations Chapter

  1. Jan 28, 2012 #1
    1. The problem statement, all variables and given/known data
    Determine an expression for the Bohr radius (a[itex]_{0}[/itex] from the following approximation. The electron moves to the nucleus to lower its potential energy,

    V(r) = -[itex]\frac{e^{2}}{r}[/itex]

    If the electron is in domain 0[itex]\leq[/itex]r[itex]\leq[/itex][itex]\bar{r}[/itex], then we may write Δp≈[itex]\frac{\hbar}{\bar{r}}[/itex], with corresponding kinetic energy [itex]\frac{\hbar^{2}}{2m\bar{r}^{2}}[/itex]. With this information estimate a[itex]_{0}[/itex] (Bohr radius) by minimizing the total energy.

    2. Relevant equations

    Uncertainty ΔpΔx≈[itex]\frac{\hbar}{2}[/itex]
    [itex]\bar{r}[/itex] is the Mean of r -------> (I think)
    [itex]\hat{H}[/itex][itex]\varphi_{n}[/itex]=E[itex]_{n}[/itex][itex]\varphi_{n}[/itex] or with x-hat and x
    For an electron moving with momentum p 2[itex]\pi[/itex]rp=nh
    E = K.E + V(x)
    [[itex]\hat{A}[/itex], [itex]\hat{B}[/itex]]=[itex]\hat{A}[/itex][itex]\hat{B}[/itex]-[itex]\hat{B}[/itex][itex]\hat{A}[/itex]
    [[itex]\hat{A}[/itex], [itex]\hat{B}[/itex]]=0 if they they commute
    [[itex]\hat{x}[/itex], [itex]\hat{p}[/itex]]=i[itex]\hbar[/itex]
    And I have no idea where this equation came from but I found it in the book and used it for my solution:
    [itex]\frac{e^{2}}{r}[/itex]=[itex]\frac{p^{2}_{\vartheta}}{mr^{2}}[/itex]=[itex]\frac{n^{2}\hbar^{2}}{mr^{2}}[/itex] where e is the electric charge of the electron and p[itex]_{\vartheta}[/itex]=mr[itex]^{2}[/itex][itex]\dot{\vartheta}[/itex]

    3. The attempt at a solution

    E = [itex]\frac{p^{2}_{\vartheta}}{2m\bar{r}^{2}}[/itex]-[itex]\frac{e^{2}}{\bar{r}}[/itex] .... i
    From the "Centripetal Condition" (The equation I got from the book) (r)[itex]\frac{e^{2}}{r^{2}}[/itex]=(r)[itex]\frac{p^{2}_{\vartheta}}{mr^{3}}[/itex]=[itex]\frac{p^{2}_{\vartheta}}{m\bar{r^{2}}}[/itex] .... ii

    sub ii into i

    E = -[itex]\frac{p^{2}_{\vartheta}}{2m\bar{r}^{2}}[/itex] ..... iii

    Now from Centripetal Condition:
    r=[itex]\frac{n^{2}\hbar^{2}}{me^{2}}[/itex] where p[itex]_{\vartheta}[/itex]=n[itex]\hbar[/itex] ..... iv

    Put iv into iii

    E = -[itex]\frac{p^{2}_{\vartheta}}{2m}[/itex]([itex]\frac{me^{2}}{n^{2}\hbar^{2}}[/itex])[itex]^{2}[/itex]=-[itex]\frac{me^{4}}{2n^{2}\hbar^{2}}[/itex] and let ℝ=[itex]\frac{m_{e}e^{4}}{2\hbar^{2}}[/itex]=13.6eV
    Then E[itex]_{n}[/itex]=[itex]\frac{ℝ}{n^{2}}[/itex]

    So for n=1, which is the lowest possible state of potential energy, E[itex]_{1}[/itex]=13.6eV

    and for n=1 from equation iv

    r[itex]_{1}[/itex]=a[itex]_{0}[/itex]=[itex]\frac{n^{2}\hbar^{2}}{me^{2}}[/itex]=5.24x10[itex]^{-9}[/itex]cm=Bohr's radius

    The problem is this was too straightforward and simple. I did not correctly use the mean value of [itex]\bar{r}[/itex], because it is a function and I am to find an expression for the Bohr radius.
    I used very little of the information provided.
    Also I just assumed the lowest energy state, whereas the question says that the electron just goes to a lower energy state.
    Also this question is in the chapter where they define the commutator relationships and compatible operators, and really the whole chapter is about that. I find it fishy the solution I provided doesn't use any of that chapters material.

    I need a new approach to this problem and am out of ideas. Please help.
    Last edited: Jan 28, 2012
  2. jcsd
  3. Jan 29, 2012 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Re: Expression for Bohr Radius using Quantum Mechanics; In Commutator Relations Chapt

    The idea here is that the electron is confined to a region of size ##\bar{r}##, so applying the uncertainty principle gives you an estimate of the magnitude of its momentum, namely ##p \approx \hbar/\bar{r}##, and therefore of its kinetic energy. The total energy of the electron is thus given as a function of ##\bar{r}## by
    $$E(\bar{r}) = \frac{\hbar^2}{2m\bar{r}^2} - \frac{e^2}{\bar{r}}.$$ The problem is simply asking you to find the value of ##\bar{r}## for which the total energy of the electron is minimized.
  4. Jan 29, 2012 #3
    Re: Expression for Bohr Radius using Quantum Mechanics; In Commutator Relations Chapt

    I see it now, thank you!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Similar Threads for Expression Bohr Radius
How to express velocity gradient in cylindrical coordinates?