A photon on a hydrogen atom excited an electron from #n = 1# to make a transition to #n=3#. What is the wavelength (in run) pertaining to this transition and was the transition emitting or absorbing in nature?

1 Answer
Dec 8, 2015

Answer:

#=>lambda=102.6nm#

Energy is absorbed.

Explanation:

In order to solve this question we will use the Bohr Model.

The change in energy according to Bohr can be calculated by:

#DeltaE=-2.178xx10^(-18)Z^(2)J(1/(n_("final")^2)-1/(n_("initial")^2))#

Here, the initial level is #n_("initial")=1# and the final level is #n_("final")=3#. Since this is hydrogen, then #Z=1#.

Thus, #DeltaE=-2.178xx10^(-18)(1)^(2)J(1/(9)-1/(1))=1.936xx10^(-18)J#

This energy can also be given as #DeltaE=hnu=hc/(lambda)#

#=>lambda=hc/(DeltaE)=(6.626xx10^(-34)xx2.998xx10^8)/(1.936xx10^(-18))=1.026xx10^(-7)m#

#=>lambda=102.6nm#

Since the electron is moving from a lower energy level to a higher energy level, the energy is absorbed.

Moreover, we can say that the energy is absorbed because the sign of #DeltaE# is positive (#DeltaE#>0).