What is ΔE for the transition of an electron from n = 7 to n = 4 in a Bohr hydrogen atom? What is the frequency, v, of the spectral line produced?

1 Answer
Jun 12, 2016

Answer:

#(a)#

#9.16xx10^(-20)" ""J"#

#(b)#

#1.38xx10^(14)" ""s"^(-1)#

Explanation:

#(a)#

The Bohr Model for the energy of an electron in a hydrogen atom gives us:

#E=(-13.6)/(n^2)" ""eV"#

Where #n# is the principle quantum number.

hyperphysics.phy-astr.gsu.edu

For the transition #7rarr4# the difference in energy is given by:

#DeltaE=-13.6/7^2-[-13.6/4^2]" ""eV"#

#DeltaE=13.6/16-13.6/49" ""eV"#

#DeltaE=0.85-0.277=0.572" ""eV"#

To convert this to Joules we multiply by the electronic charge:

#DeltaE=0.572xx1.6xx10^(-19)=9.16xx10^(-20)" ""J"#

#(b)#

To find the frequency of the emitted photon we use the Planck expression:

#DeltaE=hnu#

#:.nu=(DeltaE)/h=(9.16xx10^(-20))/(6.626xx10^(-34))" ""s"^(-1)#

#nu=1.38xx10^(14)" ""s"^(-1)#