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The energy transition will be equal to #1.55 * 10^(-19)"J"#.
So, you know your energy levels to be n = 5 and n = 3. Rydberg's equation will allow you calculate the wavelength of the photon emitted by the electron during this transition
#1/(lamda) = R * (1/n_("final")^(2) - 1/n_("initial")^(2))#, where
#lamda# - the wavelength of the emitted photon;
#R# - Rydberg's constant - #1.0974 * 10^(7)"m"^(-1)#;
#n_("final")# - the final energy level - in your case equal to 3;
#n_("initial")# - the initial energy level - in your case equal to 5.
So, you've got all you need to solve for #lamda#, so
#1/(lamda) = 1.0974 * 10^(7)"m"^(-1) * (1/3^2 - 1/5^2)#
#1/(lamda) = 0.07804 * 10^(7)"m"^(-1) => lamda = 1.28 * 10^(-6)"m"#
Since #E = (hc)/(lamda)#, to calculate for the energy of this transition you'll have to multiply Rydberg's equation by #h * c#, where
#h# - Planck's constant - #6.626 * 10^(-34)"J" * "s"#
#c# - the speed of light - #"299,792,458 m/s"#
So, the transition energy for your particular transition (which is part of the Paschen Series) is
#E = (6.626 * 10^(-34)"J" * cancel("s") * "299,792,458" cancel("m/s"))/(1.28 * 10^(-6)cancel("m"))#
#E = 1.55 * 10^(-19)"J"#
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