Does a 15 nm Photon Ionize a Ground State Hydrogen Atom?

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A 15 nm photon can ionize a ground state hydrogen atom because its energy, calculated to be approximately 82.7 eV, exceeds the ionization energy of 13.6 eV. The kinetic energy of the electron after ionization would be 69.1 eV, derived from the difference between the photon energy and the ionization energy. The calculations involve using Planck's constant and the speed of light to find the energy from the wavelength. If the hydrogen atom were in an excited state, the ionization energy would be lower, requiring only 3.4 eV to ionize. Understanding these principles is crucial for solving similar quantum mechanics problems.
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This is the exercise:
Suppose that an hydrogen atom in its ground state absorbs a photon whose wavelength is 15 nm. Will the atom be ionized? If so, what will be the kinetic energy of the electron when it
gets far away from its atom of origin?

i have no idea how to procede! Someone can help me? Thank you!
 
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first , what is the ionization energy? it is the minimm energy needed to remove an electron from an atom, and in your case you have a hydrogen atom which has an ionization energy of 13.6ev .. hint: you are given the wavelengeth of the photon find the energy from there then compare it with 13.6ev , if it is less then the atom won't be ionized .. if it is more then it will be ionized .. from there if it will be ionized, can you find out what will be the kinetic energy? ..
 
thanks for the hint. Well E=h*c/w =1.32E-17 J => 82.7 eV
So... Yes it will be ionized, and the kinetic energy should be 82.7-13.6 = 69.1 eV.
Is this right?

PS=don't know if this forum have an uniform use of the constant, in my calcolous:
h=planck costant, c=Speed of light and w = wavelenght
 
:) that seems logical enough and I believe that you got the right answer .. and welcome to the physics forum ..

but one thing I would like to add here, the problem you have would change, if for example the hydrogen atom wasnt in the ground state , if it was in the first excited state (second state) then you will need an energy of 3.4ev to ionize it ( since in this state it will have an energy = -3.4 ev, E2 = -13.6/4 ev )
 
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