CHEM
posted by K on .
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?
Enter your answer numerically in joules.
answer= delta E= _____J
what equation should I use for to answer Part B & how do I plug in to get the correct answer?

How did you get the energy for N = 4? Use the same equation but plug in 1 for N = 1 (that's the ground state). Now subtract the energy in the two states to get the amount (the difference) relesed when the electron falls from n = 4 to n = 1.

I used: E = (Rhc)/n^2
R= 1.097*10^7 m^1
h= 6.6261*10^34 J s
c= 3.00*10^8 m s^1
E= (1.097*10^7)(6.6261*10^34)(3.00*10^8)/16
E= 1.36*10^19 J
so I just use final minus initial? How so?
when I do the same as above...but put in 1squared=1, rather than 4squared=16..
I get: 2.18064951*10^18
then should the answer be: 2.18065*10^18  1.36*10^19? = 2.04*10^18?? 
guess it was correct... the real answer, rounded though, came to be: 2.05*10^18J
thank you. 
Using 2.180 x 10^18 I found 2.044 x 10^18 J for the difference in energy levels.

A microwave oven operates at 2.40 . What is the wavelength of the radiation produced by this appliance?

.125m or 125000000nm

Energy is an absolute value, the answer should be 2.05*10^18J, instead of 2.05*10^18J.

What is the wavelength lambda of the photon that has been released in Part B?

E=hc/Lambda
2.05*10^18=(6.63*10^34)(3.00*10^8)/x
x=9.72*10^8 meters