# An electron changes from an #n = 2# to an #n = 6# energy state. What is the energy of the photon in joules?

##### 1 Answer

#### Answer:

#### Explanation:

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

A good starting point here will be to calculate the energy of the photon *emitted* when the electron falls from **Rydberg equation**.

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

Here

#lamda# si thewavelengthof the emittted photon#R# is theRydberg 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 *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 theenergyof the photon#h# isPlanck's constant, equal to#6.626 * 10^(-34)color(white)(.)"J s"# #c# is thespeed of lightin 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