# Question #5da42

##### 1 Answer

Here's what I got.

#### Explanation:

Right from the start, you should know that you can use the **Rydberg equation** to find the **wavelength** of the photon emitted when an electron in a hydrogen atom falls from

#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)#

Rearrange the equation to solve for

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

and plug in your values to find

#lamda = 1/(1.097 * 10^(7)color(white)(.)"m"^(-1)) * (4^2 * 2^2)/(4^2 - 2^2)#

#color(darkgreen)(ul(color(black)(lamda = 4.86 * 10^(-7)color(white)(.)"m")))#

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

To find the **frequency** of the photon, use the fact that the wavelength of the photon and its frequency have an **inverse relationship** described by the equation

#lamda * nu = c#

Here

#nu# is thefrequencyof the photon#c# is thespeed of lightin a vacuum, usually given as#3 * 10^8# #"m s"^(-1)#

You will end up with

#nu = c/(lamda)#

#nu = (3 * 10^8 color(red)(cancel(color(black)("m"))) "s"^(-1))/(4.86 * 10^(-7)color(red)(cancel(color(black)("m"))))#

#color(darkgreen)(ul(color(black)(nu = 6.17 * 10^(14)color(white)(.)"s"^(-1))))#

Finally, to find the **energy** of the photon, use the **Planck - Einstein relation**

#E = h * nu#

Here

#E# is theenergyof the photon#h# isPlanck's constant, equal to#6.626 * 10^(-34)# #"J s"#

Plug in your values to find

#E = 6.626 * 10^(-34)color(white)(.)"J" color(red)(cancel(color(black)("s"))) * 6.17 * 10^(14)color(red)(cancel(color(black)("s"^(-1))))#

#color(darkgreen)(ul(color(black)(E = 4.09 * 10^(-19)color(white)(.)"J")))#

As a final note, the **Balmer series**.