# Question #ad4c9

##### 1 Answer

#### Answer:

#### Explanation:

The idea here is that the **difference in energy** between the initial energy level of the electron and the final energy level of the electron will give you the **wavelength** of the photon emitted during this transition.

In your case, you know that the electron falls from the sixth energy level,

#"410 nm" = 410 color(red)(cancel(color(black)("nm"))) * "1 m"/(10^9color(red)(cancel(color(black)("nm")))) = 4.10 * 10^(-7)color(white)(.)"m"#

is emitted. As you know, the **Rydberg equation** allows you to use the wavelength of the emitted photon, let's say

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

Here

#R# is theRydberg constant, equal to#1.097 * 10^(7)# #"m"^(-1)# #n_i# is theinitial energy levelof the electron#n_f# is thefinal energy levelof the electron

Now, your goal here is to figure out the value of the final energy level, so rearrange the Rydberg equation to solve for

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

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

This is equivalent to

#n_f^2 * (lamda *R + n_i^2) = lamda * R * n_i^2 #

which gets you

#n_f = sqrt( (lamda * R * n_i^2)/(lamda * R + n_i^2))#

Plug in your values to find

#n_f = sqrt( (4.10 * color(blue)(cancel(color(black)(10^(-7))))color(red)(cancel(color(black)("m"))) * 1.097 * color(blue)(cancel(color(black)(10^7)))color(red)(cancel(color(black)("m"^(-1)))) * 6^2)/(4.10 * color(blue)(cancel(color(black)(10^(-7))))color(red)(cancel(color(black)("m"))) * 1.097 * color(blue)(cancel(color(black)(10^7)))color(red)(cancel(color(black)("m"^(-1)))) + 6^2))#

#color(darkgreen)(ul(color(black)(n_f = 1.9995 ~~ 2)))#

Therefore, you can say that the electron will fall to the second energy level, **Balmer series**.

To double-check the result, you can use the **Balmer formula** to confirm that the electron is *initially located* on the sixth energy level.

#lamda = B * n_i^2/(n_i^2 - color(blue)(2)^2)#

Here

#B# is a constant that is#~~ 3.6451 * 10^(-7)color(white)(.)"m"# #n_i# is theinitial energy levelof the electron#color(blue)(2)# is thefinal energy level for the Balmer series,#n_f = 2#

This time, you will have

#n^2 * (lamda - B) = 4 * lamda#

#n = sqrt( (4 * lamda)/(lamda - B))#

Plug in your values to find

#n_i = 6.0043 ~~ 6#

which confirms that you're dealing with the