How would you calculate the energy released as a photon drops from n=3 to the ground state?

1 Answer
Jasmine
May 13, 2017

#DeltaE# =#-1.936×10^-18J#

Explanation:

#DeltaE# = #K# #(1/n_1^2-1/n_2^2)#
Where #n_1# =#1# and #n_2# =3, since the photon falls from excited to ground state the energy is released.

#DeltaE# = #K# #(1/1^2-1/3^2)#
#DeltaE# = #K# #(1/1-1/9)#
But #K# = #-2.178×10^-18J#, a constant value for calculating energy of an electron in H atom
#DeltaE# = #-2.178×10^-18##(1/1-1/9)#
#DeltaE# =#-1.936×10^_18J#
The answer is for H-atom.You didn't mentioned about the atom for which you are asking for?

Related questions
See all questions in Excited States and Ground States
Impact of this question
5227 views around the world
You can reuse this answer
Creative Commons License