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Potential Energy of hydrogen atom in ground state is 0, find energy in 1st exc state

  1. Nov 11, 2012 #1
    1. The problem statement, all variables and given/known data
    What is the energy of H atom in the first excited state if the potential energy in the ground state is taken to be 0 ?

    2. Relevant equations
    Usually the energy of H atom in the ground state is -13.6eV
    and in the 1st excited state is -10.2eV
    E(n) = πme2/8ε2h2
    Bohr's radius = 0.529 angstrom

    The answer provided is : 23.8eV

    3. The attempt at a solution

    So I went with how the energy of the electron is derived from ground up
    Knowing that
    Total energy = potential energy + Kinetic energy
    potential energy due to the positive charge on nucleus and negative on electron
    so if the potential energy is taken to be 0 then the only energy left is the kinetic one so
    the energy would just be
    1/2 mvv
    then I substituted
    v = e2/2∏hn

    then I used the formula
    ΔE = 13.6(3/4)

    but everything is now messed up and I ended up with a gross looking equation , I know something doesn't makes sense above may be nothing does .

    Thanks in advance :

    P.S : the answer is 23.8 and 13.6 + 10.2 = 23.8
    but how can I possibly do this ?
     
  2. jcsd
  3. Nov 11, 2012 #2

    TSny

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    Re: Potential Energy of hydrogen atom in ground state is 0, find energy in 1st exc st

    Check the value of the energy of the 1st excited state.
     
  4. Nov 11, 2012 #3
    Re: Potential Energy of hydrogen atom in ground state is 0, find energy in 1st exc st

    oops! 10.2eV is required to excite the electron from the ground to the first excited state.
     
  5. Nov 11, 2012 #4

    TSny

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    Gold Member

    Re: Potential Energy of hydrogen atom in ground state is 0, find energy in 1st exc st

    These don't look correct to me.
     
  6. Nov 11, 2012 #5
    Yeah it isn't
    In the E e^2 should be e^4

    And in velocity pi should be replaced by epsilon

    I posted this from my phone .My bad
     
  7. Nov 11, 2012 #6
    Re: Potential Energy of hydrogen atom in ground state is 0, find energy in 1st exc st

    Ok,to find the potential energy of the hydrogen atom in the first excited state excluding fine structures [itex]\Longrightarrow[/itex]

    To find the energy levels of the quantum states of the hydrogen atom,we must solve Schrodinger's equation,with equation U = -ke²/r substituted for U in that equation.Solving that equation reveals that the energies of the electron's quantum states are given by,
    E[itex]_{n}[/itex] = -m[itex]_{e}[/itex]e[itex]^{4}[/itex]/8ε[itex]_{o}²[/itex]h²n² which leads us to [itex]\Longrightarrow[/itex]

    -13.6 eV/n[itex]^{2}[/itex] , for n=1,2,3,...,

    The value of 13.6 eV is called the Rydberg constant and can be found from the Bohr model.

    For n = 2 (first excited state) we get E[itex]_{2}[/itex] = -3.4 eV
    It means that the energy required to excite an electron in hydrogen atom to its first excited state,is an energy equal to E[itex]_{2}[/itex] - E[itex]_{1}[/itex] = -3.4 eV-(-13.6)eV
    = 10.2 eV

    → E[itex]_{n}[/itex] = -mZ[itex]^{2}[/itex]e[itex]^{4}[/itex]/8ε[itex]_{0}[/itex][itex]^{2}[/itex]n[itex]^{2}[/itex]h[itex]^{2}[/itex]
    → For n=2,Z=2
    → E = -8.5 × 10[itex]^{19}[/itex] eV
     
  8. Nov 11, 2012 #7
    Re: Potential Energy of hydrogen atom in ground state is 0, find energy in 1st exc st

    Ok,to find the potential energy of the hydrogen atom in the first excited state excluding fine structures [itex]\Longrightarrow[/itex]

    To find the energy levels of the quantum states of the hydrogen atom,we must solve Schrodinger's equation,with equation U = -ke²/r substituted for U in that equation.Solving that equation reveals that the energies of the electron's quantum states are given by,
    E[itex]_{n}[/itex] = -m[itex]_{e}[/itex]Z[itex]^{2}[/itex]e[itex]^{4}[/itex]/8ε[itex]_{o}²[/itex]h²n² which leads us to [itex]\Longrightarrow[/itex]

    -13.6 eV/n[itex]^{2}[/itex] , for n=1,2,3,...,

    The value of 13.6 eV is called the Rydberg constant and can be found from the Bohr model.

    For n = 2 (first excited state) we get E[itex]_{2}[/itex] = -3.4 eV
    It means that the energy required to excite an electron in hydrogen atom to its first excited state,is an energy equal to E[itex]_{2}[/itex] - E[itex]_{1}[/itex] = -3.4 eV-(-13.6)eV
    = 10.2 eV

    → E[itex]_{n}[/itex] = -m[itex]_{e}[/itex]Z[itex]^{2}[/itex]e[itex]^{4}[/itex]/8ε[itex]_{0}[/itex][itex]^{2}[/itex]n[itex]^{2}[/itex]h[itex]^{2}[/itex]
    → For n=2,Z=2
    → E = -8.5 eV

    I have done this problem including Z therefore my answer is -8.5 eV.Now adding them 10.2 eV + (-8.5)eV
    = 1.7 eV

    But according to your answer,it looks different! :surprised
     
    Last edited: Nov 11, 2012
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