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 lower-energy 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?

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