One of the lines in the emission spectrum of the hydrogen atom has a wavelength of 102.6 nm. How would you identify the value of n-initial for the transition giving rise to this emission?

1 Answer
Dec 3, 2015

Answer:

#n_("initial")=0.73" "#???!!!!!

Explanation:

You can use Bohr model where the change on energy is calculated by:

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

#DeltaE# can be calculated from the wavelength by:

#DeltaE=(hc)/lambda=(6.626xx10^(-34)xx2.998xx10^8)/(102.6xx10^(-9))=1.94xx10^(-18)J#

Therefore,
#=>1.94xx10^(-18)=-2.178xx10^(-18)xx(1)^2xx((1)/((1)^2) - 1/((n_("initial"))^2))#

#=>1.94xx10^(-18)=-2.178xx10^(-18)+(2.178xx10^(-18))/((n_("initial"))^2)#

Solve for #n_("initial")=0.73# This is weird. I think there is a problem with the wavelength given.

Anyways, this is how I would solve it.

I hope you find this helpful!