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Note on problem solving.
A handy way to calculate photon energies and wavelengths is as follows: We know
E = hc/lambda and  = hc/E
for photons. The quantity hc may be expressed as 1240 eV-nm. Thus, a 2.0-eV photon as a wavelength
lambda = (1240 eV-nm)/(2.0 eV) = 620 nm
A 1.4-nm x-ray has an energy of
E = (1240 eV-nm)/(1.4 nm) = 886 eV
This avoids the messy and cumbersome use of exponents when using Joules.
1. (A) Looking at Figure 42-1, find the energy of a photon emitted from an n = 2 to n = 1 transition, expressing the answer in eV. Now find the wavelength. Is this photon in the infrared, the visible, or the ultraviolet part of the spectrum? ( 122 nm )
(B) Photons emitted from the Balmer series of hydrogen are due to transitions of n values 3 to 2, 4 to 2, 5 to 2, and so on. Draw arrows on the diagram below, representing the first three transitions of this series.© BrainMass Inc. brainmass.com October 24, 2018, 7:41 pm ad1c9bdddf
This solution provides step by step calculations and explanations for finding the energy of a photon and drawing transitions from a Balmer series.
Bohr Theory and Balmer Series
One of the lines in the Balmer series of the hydrogen atom emission spectrum is at 397nm. It results from a transition from an upper level energy level to n=2. What is the principal quantum number of the upper level?View Full Posting Details