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?
E=hc/Lambda
2.05*10^-18=(6.63*10^-34)(3.00*10^8)/x
x=9.72*10^-8 meters
Energy is an absolute value, the answer should be 2.05*10^-18J, instead of -2.05*10^-18J.
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.