Is energy emitted or absorbed when the electronic transition from n 3 to n = 6 occurs in hydrogen

The question wants you to determine the energy that the incoming photon must have in order to allow the electron that absorbs it to jump from #n_i = 2# to #n_f = 6#.

A good starting point here will be to calculate the energy of the photon emitted when the electron falls from #n_i = 6# to #n_f = 2# by using the Rydberg equation.

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

Here

• #lamda# si the wavelength of the emittted photon
• #R# is the Rydberg constant, equal to #1.097 * 10^(7)# #"m"^(-1)#

Plug in your values to find

#1/lamda = 1.097 * 10^7color(white)(.)"m"^(-1) * (1/2^2 - 1/6^2)#

#1/lamda = 2.4378 * 10^6 color(white)(.)"m"^(-1)#

This means that you have

#lamda = 4.10 * 10^(-7)color(white)(.)"m"#

So, you know that when an electron falls from #n_i = 6# to #n_f = 2#, a photon of wavelength #"410 nm"# is emitted. This implies that in order for the electron to jump from #n_i = 2# to #n_f = 6#, it must absorb a photon of the same wavelength.

To find the energy of this photon, you can use the Planck - Einstein relation, which looks like this

#E = h * c/lamda#

Here

• #E# is the energy of the photon
• #h# is Planck's constant, equal to #6.626 * 10^(-34)color(white)(.)"J s"#
• #c# is the speed of light in a vacuum, usually given as #3 * 10^8 color(white)(.)"m s"^(-1)#

As you can see, this equation shows you that the energy of the photon is inversely proportional to its wavelength, which, of course, implies that it is directly proportional to its frequency.

Plug in the wavelength of the photon in meters to find its energy

#E = 6.626 * 10^(-34) color(white)(.)"J" color(red)(cancel(color(black)("s"))) * (3 * 10^8 color(red)(cancel(color(black)("m"))) color(red)(cancel(color(black)("s"^(-1)))))/(4.10 * 10^(-7) color(red)(cancel(color(black)("m"))))#

#color(darkgreen)(ul(color(black)(E = 4.85 * 10^(-19)color(white)(.)"J")))#

I'll leave the answer rounded to three sig figs.

So, you can say that in a hydrogen atom, an electron located on #n_i = 2# that absorbs a photon of energy #4.85 * 10^(-19)# #"J"# can make the jump to #n_f = 6#.

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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