Posted by sam on Monday, October 25, 2010 at 5:41pm.
The electron from a hydrogen atom drops from an excited state into the ground state. When an electron drops into a lowerenergy orbital, energy is released in the form of electromagnetic radiation.
Part A: How much energy does the electron have initially in the n=4 excited state?
answer I got = 1.36*10^19 J
Part B: If the electron from Part A now drops to the ground state, how much energy is released?
the answer i got was 2.04*10^18 J
but i don't understand part :
What is the wavelength lambda of the photon that has been released in Part B?
can someone please explain this one to me, thank you

chemistry  DrBob222, Monday, October 25, 2010 at 6:25pm
I didn't check your answers to A and B.
For the part you don't understand, subtract the energy of the electron in the two levels; that is the energy released. Then substitute into E = hc/wavelength and solve for wavelength.The unit for wavelength in this equation is meter. 
chemistry  helpplease, Tuesday, October 26, 2010 at 7:28pm
When an electron moves from the n=2 to n=4 state, Does it gain or lose energy? In a more to n=4, is a photon emitted or absorbed?