# How much energy is released when #"1 mol"# of hydrogen atoms transition from #n=3# to #n=2#?

##### 1 Answer

Here's what I got.

#### Explanation:

The first thing that you need to do here is to figure out the **wavelength** of the photon emitted by the electron as it's making the

To do that, you need to use the **Rydberg equation**, which looks like this

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

Here

#lamda# is thewavelengthof the photon#R# is theRydberg constant, equal to#1.097 * 10^(7)# #"m"^(-1)# #n_I# is theinitial energy levelof the transition#n_f# is thefinal energy levelof the transition

Plug in your values to find

#1/(lamda) = 1.097 * 10^(7)color(white)(.)"m"^(-1) * (1/2^2 - 1/3^2)#

#1/(lamda) = 1.097 * 10^(7)color(white)(.)"m"^(-1) * ((9 - 4)/(9 * 4))#

Rearrange to solve for

#lamda = 36/(1.097 * 10^(7) * 5)color(white)(.)"m"#

#lamda = 6.563 * 10^(-7)color(white)(.)"m"#

The wavelength of the emitted photon confirms that you're dealing with a transition that's part of the **Balmer series**.

Now, the **energy** of a photon is **directly proportional** to its *frequency*, which implies that it is **inversely proportional** to its *wavelength*.

This relationship is described by the **Planck - Einstein relation**

#E = h * c/(lamda)#

Here

#E# is theenergyof the photon#h# isPlanck's constant, equal to#6.626 * 10^(-34)# #"J s"# #c# is thespeed of lightin a vacuum, usually given as#3 * 10^8# #"m s"^(-1)#

Plug in your value to find the energy of a *single photon* that is emitted when an electron makes this transition.

#E = 6.626 * 10^(-34)"J" color(red)(cancel(color(black)("s"))) * (3 * 10^8color(red)(cancel(color(black)("m"))) color(red)(cancel(color(black)("s"^(-1)))))/(6.563 * 10^(-7) color(red)(cancel(color(black)("m"))))#

#E = 3.029 * 10^(-19)# #"J"#

So, you need the **energy** of the photon because it is **equal** to the difference between the two energy levels of the transition.

In other words, the difference between the energy of the

#DeltaE_ (3 -> 2) = 3.029 * 10^(-19)# #"J"#

This value corresponds to the energy of **atom** of hydrogen.

To find the energy difference for **moles** of hydrogen atoms, you need to find the total energy of **moles** of photons. Your tool of choice here will be **Avogadro's constant**, which tells you that **mole** of photons will contain **photons**.

#6.022 * 10^(23) color(red)(cancel(color(black)("photons"))) * (3.029 * 10^(-19)color(white)(.)"J")/(1color(red)(cancel(color(black)("photon")))) = color(darkgreen)(ul(color(black)(1.82 * 10^(5)color(white)(.)"J")))#

The answer is rounded to three **sig figs**. If you want, you can express the answer in *kilojoules* to get