Kinetic energy of an ionized electron

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Ultraviolet light at a wavelength of 60.0 nm ionizes hydrogen atoms, releasing electrons. The kinetic energy of these freed electrons is determined by the energy of the incoming photon minus the ionization energy of hydrogen. The energy of the photon can be calculated using the equation E = hc/λ, where h is Planck's constant and c is the speed of light. The ionization energy for hydrogen in its ground state is approximately 13.6 eV. The remaining energy after ionization contributes to the kinetic energy of the electrons.
Linus Pauling
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1. Ultraviolet light with a wavelength of 60.0 nm shines on a gas of hydrogen atoms in their ground states. Some of the atoms are ionized by the light. What is the kinetic energy of the electrons that are freed in this process?



2. En = n2h2/8mL2



3. To get L, I solved lambda = 2L/n with n = 1, obtaining L = 3*10-8 m. Plugging to the En equation, using n =1 and h = 4.14*10-15 eVs, I obtain an incorrect answer. I tried using both the Js and EvS numbers for h, but I was wrong either way...
 
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The energy of a hydrogen atom is not the same as the energy of an infinite square well...

But that doesn't matter.

What you want to see is how much of the energy of the photon goes to ionizing the electron? (Hint: consider what the energy of an electron in the ground state of hydrogen is) Therefore, how much energy is left to give to the kinetic energy of the electron?
 

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