How much energy is emitted when an electron moves from the n=4 state of a hydrogen atom to the n=2 state?

1 Answer
Apr 30, 2016

I found: #DeltaE=4.09xx10^-19J#

Explanation:

I would use the energy of the electron in each state as:
#E_n=-(13.6eV)/n^2=-(2.18xx10^-18J)/n^2#
so we get:
#n=2#
#E_2=-(2.18xx10^-18J)/2^2=-5.45xx10^-19J#
#n=4#
#E_4=-(2.18xx10^-18J)/4^2=-1.36xx10^-19J#

#DeltaE=E_4-E_2=-1.36xx10^-19J+5.45xx10^-19J=4.09xx10^-19J#

Corresponding to a difference of #2.55eV#.