CHEM

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?
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?

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  1. 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.

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  2. 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??

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  3. guess it was correct... the real answer, rounded though, came to be: -2.05*10^-18J

    thank you.

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  4. Using 2.180 x 10^-18 I found 2.044 x 10^-18 J for the difference in energy levels.

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

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  6. .125m or 125000000nm

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

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

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  9. E=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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  10. c is used for sun tan

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