An electron in a hydrogen atom drops from energy level n=5 to n=3. What is the energy transition using the Rydberg equation?

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Mar 23, 2015

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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Michael Share
Mar 24, 2015

A photon of energy #1.531xx10^(-19)"J"# will be emitted.

The Rydberg expression for an electronic transition in the hydrogen atom is:

#1/lambda=R[(1)/n_1^(2)-(1)/n_2^(2)]#

#lambda="wavelength"#

#R=1.097xx10^(7)m^(-1)#

#n_1=3#

#n_2=5#

So:

#1/lambda=R(1/9-1/25)#

#1/lambda=R(0.071)=1.097xx10^(7)xx0.071=7.789xx10^(5)m^(-1)#

The energy of the photon given by this transition is given by:

#E=hf#

Since #c=flambda#

#E=(hc)/(lambda)#

#h# is the Planck Constant = #6.63xx10^(-34)"Js"#

#c# = the speed of light =#3xx10^(8)"m/s"#

So:

#E=6.63xx10^(-34)xx3xx10^(8)xx7.789xx10^(5)"J"#

#E=1.531xx10^(-19)"J"#

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