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Ground state energy for hydrogen-like “atom”

  1. Mar 29, 2015 #1
    1. The problem statement, all variables and given/known data
    For a hydrogen-like “atom” (e.g., He+ ion), with nuclear charge Ze, it is claimed that the the ground-state wavefunction is spherically symmetric and is given by ψ(r)=Aexp(−αr) , where A and α are constant. (a) Determine the normalization constant A in terms of α. (b) Determine α by making use of (Eq.3) in the lecture note (Ch.9 pg 4). Note that you need to make some change in the equation before applying it to this system. Hence, find the ground state energy.

    The mentioned equation here is attached below. The R in the equation refers to ψ(r) and l is the orbital quantum number



    2. Relevant equations


    3. The attempt at a solution

    I can finish part a) but for part b), i substitute ψ(r)=Aexp(−αr) and try to solve for α. It comes up with the following result and im stuck. please help!
     

    Attached Files:

    Last edited: Mar 29, 2015
  2. jcsd
  3. Mar 29, 2015 #2

    mfb

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    It is hard to read your handwriting, can you write it here (as text or with LaTeX)?
    You know l for the ground state, that allows to simplify the equation significantly.
     
  4. Mar 29, 2015 #3

    TSny

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    Hi, Andy. Welcome to PF!

    Looks like you dropped a factor of ##\hbar## in going from the first to second line.

    Look at your 6th line (before you solved for ##\alpha##). Note that it must be satisfied for all values of ##r## even though ##\alpha## is a constant.
     
    Last edited: Mar 29, 2015
  5. Mar 29, 2015 #4
    I'm sorry my hand-writing is a bit messy. I will try to learn how to use LaTex
     
  6. Mar 29, 2015 #5
    I understand that ##\alpha## is a constant and when i solve it, it somehow seems like to depend on r. And when i check my calculations, it seems to be ok except that ##\hbar## part. What shall i do? Can i just randomly let r be any number i want?
     
  7. Mar 29, 2015 #6

    mfb

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    For the correct α, the equation does not depend on r any more. Or, in other words, the derivative of "something=0" with respect to r will be zero.
     
  8. Mar 29, 2015 #7

    TSny

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    Right. ##\alpha## can't depend on ##r##. This fact actually determines the value of ##\alpha##. By examining your 6th equation, can you see a way for the two terms that depend on ##r## to cancel out?
     
  9. Mar 29, 2015 #8
    since ##\alpha##. doesnt depends on ##r## and ##E## doesnt depend on ##r## too, the only way to cancel them is to put those two terms to be zero when added up and then i can solve for ##\alpha##.?
     
  10. Mar 29, 2015 #9

    TSny

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    Yes, I think that's right. Combine the two terms that contain ##r## and see if you can make the combined term vanish.
     
  11. Mar 29, 2015 #10
    Yes. I can find the answer now. Thanks all of you!:smile::smile::smile:
     
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