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!

Electric Potential and Ionization Energy of Bohr Hydrogen

  1. Feb 2, 2016 #1
    I was studying for (first year) physics class and was playing around with the Bohr Model of Hydrogen. I tried calculating the electric potential at the Bohr radius r =5.29e-11 m, where [itex] V = \frac{e}{4 \pi \epsilon_0 r} [/itex] (from the point-charge formula for electric potential) and I got 27.19 J/C, but I am confused because the ionization energy of Hydrogen is 13.6eV, and doesn't that mean that the electron must be sent through 13.6 volts of electric potential?

    I tried to calculate the potential and kinetic energies of the electron, [itex] U = qV =-eV [/itex] and [itex] {F_c} = \frac{mv^2}{r} [/itex] and [itex] F = qE = \frac{qV}{r}=\frac{+eV}{r}[/itex]

    Therefore [itex] \frac{mv^2}{r}=\frac{qV}{r} →\frac{1}{2}mv^2 = \frac{+eV}{2}[/itex]


    So then this would seem to suggest to me that the net energy of the electron is simply [itex]K+U=\frac{1}{2} U=\frac{qV}{2}=\frac{-eV}{2}[/itex]

    Does this imply that the (work) ionization energy is simply [itex]W=-U=-(\frac{-eV}{2})=+e(13.6V)=+13.6eV[/itex]?

    Because it makes sense to me (assuming my calculations are correct) when calculating the work that the energy of an electron would just be 13.6eV at the Bohr radius, but doesn't make sense to me when it just passed through 27.19 volts of electric potential. Can someone please explain this to me?

    Thanks a lot!
  2. jcsd
  3. Feb 2, 2016 #2


    User Avatar
    Homework Helper

    The electron in the nucleus has some potential energy (negative) and some kinetic energy (positive). The total energy of the bound electron is E=U+K=-eV/2. The ionization energy is the minimum work needed to move the electron to infinity,where both the potential energy and the kinetic energy are zero, so E+W=0, W = -E = eV/2. The electron uses its own kinetic energy to climb out of the potential well, so less external work is needed.
  4. Feb 12, 2016 #3
    I think your confusion is due to wrong conceptual base of the situation in which the electron is -
    first you must think of electron in a bound state- it means some invisible rope is pulling it and rotating it in a circular orbit of radius r
    - a bound state can only exist with total energy of electron to be negative-
    even if its total energy goes to zero its not bound and can move to free state outside the grip of the nucleus- so when its free the atom is ionised and has lost the electron so can be called 'ionised' having a net +ve electronic charge.

    so the amount of energy required to unbound the electron will be the amount of energy required that is the 13.6 eV needed as ionisation energy.

    in the above explanation if you wish to calculate the work done then you write work done dw in bringing the elctron say from r to r+dr which one can write as dw= F.dr (a scalar product) .
    put in the value of F which you know and integrate the expression from r=0 to r= infinity
    and i think with this you should get to the right answer -
    try as this example to find out escape velocity of bodies from earth which are bound to earth in a similar gravitational field see whether it satisfies you.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook