The energy of the photon emitted shall equal to the difference between the electron's potential energy at the two orbits.

The #ul("Rydberg Formula")# suggests that an electron that travels (or literally "falls") from energy level #n_i# to #n_f# (#n_i>n_f#) emits a photon of wavelength #lambda# for which

#1/lambda=R(1/n_f^2-1/n_i^2)#

where the Rydberg's Constant #R=1.097 xx 10^7 color(white)(l) m^{-1}#.

In this particular scenario

Such that

#1/lambda = R(1/n_f^2-1/n_i^2)#

#color(white)(Delta "PE") = 1.097 xx 10^7 color(white)(l) m^(-1)* (1/1^2-1/3^2)#

#color(white)(Delta "PE") = 9.751*10^7 color(white)(l) m^(-1)#

#lambda = 1.026 xx 10^(-7) color(white)(l) color(red)(cancel(color(black)(m))) * (10^9 color(white)(l) nm)/(1 color(white)(l) color(red)(cancel(color(black)(m))))#

#color(white)(lambda) = 102.6 color(white)(l) nm#

**References**

"Bohr's Hydrogen Atom" , Chemistry Libretext