Consider an electron in the n = 2 energy level of a hydrogen atom. (a) What is t
ID: 533743 • Letter: C
Question
Consider an electron in the n = 2 energy level of a hydrogen atom. (a) What is the energy of the n = 2 energy level? (Give your answer to three significant figures.) J (b) How much energy is involved in the transition of the electron from n = 2 to n = 1? (Give your answer to three significant figures.) J (c) Is the energy you calculated in (b) absorbed or released during the transition? absorbed released (d) What wavelength of electromagnetic radiation is absorbed or emitted during the transition in (b)? (Give your answer to three significant figures.) m (e) Classify the type of electromagnetic radiation you determined in (d). radio wave microwave infrared visible ultraviolet X-ray gamma rayExplanation / Answer
When an electron relaxes from a higher energy level to a lower level it sheds the excess energy, a positive amount, by emitting a photon. The energy of the emitted photon is given by the Rydberg Formula. This formula is essentially the subtraction of two energy levels. It is:
E = E0 ((1/n12 )- (1/n22)), where E0 is the magnitude of the energy of an electron hydrogen atom in the first orbit (13.6 eV), n1 and n2 and initial and final orbits
Since the energy of the electron in hydrogen atom in nth orbit is = -E0 /n2
E = -13.6/4 = 3.4 eV which is equals
-3.4 x 1.6 x 10-19 = -5.440 x 10-19 J (note 1 eV = 1.6 x 10-19 J)
b. As mentioned earlier, the energy difference between n = 2 and n= 1 = (-13.6 +3.4) = 10.2 eV =
= 16.32 x 10-19 J
Where h is planks constant (6.626 x 10-34 Js)
C is velocity of light (3 x 108m/s)
l is the wavelength
Substituting values
l = (6.626 x 10-34 * 3 x 108)/ (16.32 x 10-19) = 1.21801 x 10-7 m = 121.801 nm
e this wavelength region is coming in ultraviolet region (10-370 nm)