Question #cf90a
1 Answer
Explanation:
That formula will not give you the wavelength of the emitted photon, it will give you its energy.
You would then use the Planck - Einestein equation to get the wavelength of the photon.
That value given to you ,
So, the energy of a photon emitted when an electron drops from the
#E = R_E * (1/n_1^2 - 1/n_2^2)" "# , where
In your case, the energy of the photon would be
#E = 2.18 * 10^(-18)"J" * (1/1^2 - 1/6^2)#
#E = 2.18 * 10^(-18)"J" * (1 - 1/36) = 2.12 * 10^(-18)"J"#
The Planck - Einstein equation looks like this
#E = h * nu" "# , where
You know that frequency and wavelength have an inverse relationship
#nu * lamda = c" "# , where
This means that the relationship between energy and wavelength looks like this
#E = h * c/(lamda)#
Plug in your values and rearrange to solve for
#lamda = (h * c)/E = (6.626 * 10^(-34)color(red)(cancel(color(black)("J"))) color(red)(cancel(color(black)("s"))) * 3.0 * 10^8"m" color(red)(cancel(color(black)("s"^(-1)))))/(2.12 * 10^(-18)color(red)(cancel(color(black)("J"))))#
#lamda = 9.376 * 10^(-8)"m" = "93.8 nm"#
SIDE NOTE You can incorporate the Planck - Einstein equation into the Rydberg equation
#E = R_E * (1/n_1^2 - 1/n_2^2)#
#h * c/(lamda) = R_E * (1/n_1^2 - 1/n_2^2)#
Rearrange to get
#1/(lamda) = R_E/(h * c) * (1/n_1^2 - 1/n_2^2)#
The term
#R = (2.18 * 10^(-18)color(red)(cancel(color(black)("J"))))/(6.626 * 10^(-34)color(red)(cancel(color(black)("J"))) color(red)(cancel(color(black)("s"))) * 3.0 * 10^8 "m" color(red)(cancel(color(black)("s"^(-1))))) ~~ 1.9073... * 10^(7)"m"^(-1)#