Calculate the energy of a photon emitted when an electron in a hydrogen atom relaxes from n=5 to n=3?

1 Answer
Apr 30, 2018

The energy is #1.549 × 10^"-19" color(white)(l)"J"#.

Explanation:

You can use the Rydberg formula to calculate the change in energy.

Rydberg's original formula was in terms of wavelengths, but we can rewrite it to have the units of energy.

The formula then becomes

#color(blue)(bar(ul(|color(white)(a/a) ΔE = R(1/n_text(f)^2 - 1/n_text(i)^2)color(white)(a/a)|)))" "#

where

#R = "the Rydberg constant", 2.178 × 10^"-18"color(white)(l) "J"#
#n_text(i)# and #n_text(f)# are the initial and final energy levels.

For a transition from #n = 5# to #n = 3#, we get

#ΔE = 2.178 × 10^"-18"color(white)(l) "J"(1/3^2 -1/5^2) = 2.178 × 10^"-18"color(white)(l) "J"(1/9 -1/25)#

#= 2.178 × 10^"-18"color(white)(l) "J" × ("25 - 9")/(25 × 9) = 2.178 × 10^"-18"color(white)(l) "J" × 16/225 = 1.549 × 10^"-19" color(white)(l)"J"#