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

1 Answer
May 13, 2017

Answer:

#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?