An electron in an hydrogen atom makes a transition from the ground state, #n=1#, to the excited state, #n=5#. Calculate the energy in Joules, frequency in Hertz #(1/s)#, and wavelength in #nm# of the photon?

Wavelength

You use the Rydberg formula to calculate the wavelength, λ:

#color(blue)(bar(ul(|color(white)(a/a) 1/λ = RZ^2(1/n_2^2 -1/n_1^2)color(white)(a/a)|)))" "#

where

#R =# the Rydberg constant (#1.097 × 10^7color(white)(l) "m"^"-1"#)
#Z =# the atomic number of the atom
#n_1# and #n_2# are the initial and final energy levels

In this problem,

#Z = 1#
#n_1 = 1#
#n_2 = 5#

#1/λ = 1.097 × 10^7color(white)(l) "m"^"-1" × 1^2 (1/5^2 -1/1^2) = 1.097 × 10^7color(white)(l) "m"^"-1" (1/25-1/1)#

#= 1.097 × 10^7color(white)(l) "m"^"-1" × (1 - 25)/25 = "-1.097" × 10^7color(white)(l) "m"^"-1" × 24/25 = "-1.053" × 10^7 color(white)(l)"m"^"-1"#

The negative sign shows that energy is absorbed.

#λ = 1/(1.0531 × 10^7 color(white)(l)"m"^"-1") = 9.496 × 10^"-8" "m" = "94.96 nm"#

Frequency

The formula relating frequency #f# and wavelength #λ# is

#color(blue)(bar(ul(|color(white)(a/a)fλ = c color(white)(a/a)|)))" "#

where #c# is the speed of light (#2.998 × 10^8color(white)(l) "m·s"^"-1"#).

We can rearrange the formula to get

#f = c/λ#

∴ #f = (2.998×10^8 color(red)(cancel(color(black)("m")))"·s"^"-1")/(94.96 × 10^"-9" color(red)(cancel(color(black)("m")))) = 3.157 × 10^15color(white)(l) "s"^"-1"#

The formula relating energy #E# and frequency #f# is

#color(blue)(bar(ul(|color(white)(a/a)E = hfcolor(white)(a/a)|)))" "#

where #h# is Planck's constant #6.626 × 10^"-34"color(white)(l) "J·s"#

∴ #E = 6.626 × 10^"-34"color(white)(l) "J"·color(red)(cancel(color(black)("s"))) × 3.157 × 10^"15" color(red)(cancel(color(black)("s"^"-1"))) = 2.092 × 10^"-19"color(white)(l) "J"#