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$\bullet$ (a) How much energy is needed to ionize a hydrogen atom that is in the $n=4$ state? (b) What would be the wavelength of a photon emitted by a hydrogen atom in a transition from the $n=4$ state to the $n=2$ state?

$| 487 \mathrm{nm}$

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all right. So in hydrogen, the energy associate id with the quantum number and is given by negative 13.6 TV over and squared. And in part a, we were asked how much energy is needed to, um, ionized an electron from the n equals force. They So when the ionizing electron weaken, basically think of it as being removed infinitely far from the hydrogen atom or practically infinite, in which case the energy of the electron is a zero. So basically the energy that we have to put in eyes just equal to the energy that the electron initially had, which for an an equals for quantum stay is given by 13.6 TV over four squared, and this is 0.85 BB. Now, if we have a transition between Orbital's two and four, then we need to take a difference of energies using the same expression. So the energy associate ID with the N equals to orbital is, um, 13.6 TV over one over two squared, and we take the difference with the N equals for orbital and the energy supplied, um, to the photon is given by the difference of these energies. Basically, the change in energy in the orbital's is compensated for by the mission or absorption of the photo, and this comes out to 2.55 If we want to know the wavelength, then we can set the sequel to each C over Lambda with H expressed in electron volts and solving for land gives 487 nano meters, I think.

University of Hawaii at Manoa