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March 26, 2015

March 26, 2015

Posted by **K** on Sunday, November 4, 2007 at 2:02pm.

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?

- CHEM -
**DrBob222**, Sunday, November 4, 2007 at 2:12pmHow 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.

- CHEM -
**K**, Sunday, November 4, 2007 at 2:22pmI 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??

- CHEM -
**K**, Sunday, November 4, 2007 at 2:26pmguess it was correct... the real answer, rounded though, came to be: -2.05*10^-18J

thank you.

- CHEM -
**DrBob222**, Sunday, November 4, 2007 at 2:31pmUsing 2.180 x 10^-18 I found 2.044 x 10^-18 J for the difference in energy levels.

- CHEM -
**john**, Thursday, April 3, 2008 at 3:14pmA microwave oven operates at 2.40 . What is the wavelength of the radiation produced by this appliance?

- CHEM -
**Rita**, Tuesday, March 31, 2009 at 9:02pm.125m or 125000000nm

- CHEM -
**Ramiz Alda**, Sunday, November 22, 2009 at 2:42pmEnergy is an absolute value, the answer should be 2.05*10^-18J, instead of -2.05*10^-18J.

- CHEM -
**sam**, Monday, October 25, 2010 at 5:34pmWhat is the wavelength lambda of the photon that has been released in Part B?

- CHEM -
**ZAR**, Wednesday, December 8, 2010 at 10:22amE=hc/Lambda

2.05*10^-18=(6.63*10^-34)(3.00*10^8)/x

x=9.72*10^-8 meters

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