One series of lines of the hydrogen spectrum is caused by...

One series of lines of the hydrogen spectrum is caused by emission of energy accompanying the fall of an electron from outer shells to the fourth shell. The lines can be calculated using the Balmer-Rydberg equation: $$ \frac{1}{\lambda}=R_{\infty}\left[\frac{1}{m^{2}}-\frac{1}{n^{2}}\right] $$ where $m=4, R_{\infty}=1.097 \times 10^{-2} \mathrm{~nm}^{-1}$, and $n$ is an integer greater than $4 .$ Calculate the wavelengths in nanometers and energies in kilojoules per mole of the first two lines in the series. In what region of the electromagnetic spectrum do they fall?

One series of lines of the hydrogen spectrum is caused by emission of energy accompanying the fall of an electron from outer shells to the fourth shell. The lines can be calculated using the Balmer-Rydberg equation:
$$
\frac{1}{\lambda}=R_{\infty}\left[\frac{1}{m^{2}}-\frac{1}{n^{2}}\right]
$$
where $m=4, R_{\infty}=1.097 \times 10^{-2} \mathrm{~nm}^{-1}$, and $n$ is an integer greater than $4 .$ Calculate the wavelengths in nanometers and energies in kilojoules per mole of the first two lines in the series. In what region of the electromagnetic spectrum do they fall?

Key Concepts

Rydberg Formula

The Rydberg formula relates the wavelengths of emitted or absorbed light to the transitions between energy levels in a hydrogen atom. It is expressed as 1/? = R(1/m² - 1/n²), where R is the Rydberg constant and m and n are the principal quantum numbers of the lower and higher energy levels, respectively. This formula is central in predicting and calculating the positions of spectral lines.

Electromagnetic Spectrum Regions

The electromagnetic spectrum is divided into regions such as ultraviolet, visible, and infrared, among others. Determining the region in which a spectral line falls involves calculating its wavelength and comparing it to the known boundaries between these regions, which is essential for understanding the observational characteristics and applications of the spectral line.

Photon Energy and Energy Conversion

The energy of a photon is related to its wavelength by the equation E = hc/?, where h is Planck’s constant and c is the speed of light. Calculating the energy in joules for a single photon and then converting it to kilojoules per mole (using Avogadro’s number) allows one to relate microscopic energy transitions to macroscopic amounts relevant in chemical and physical processes.

Atomic Electron Transitions

Atomic electron transitions involve an electron moving between discrete, quantized energy levels in an atom. When an electron falls from a higher energy level to a lower one, it emits a photon whose energy equals the difference between these levels. This fundamental phenomenon underlies the formation of spectral lines in the atomic emission spectrum.

Spectral Series Classification

Spectral series in hydrogen are defined by the electron’s final energy level in a transition. Common series include Lyman (m=1), Balmer (m=2), Paschen (m=3), and Brackett (m=4), among others. Each series is associated with a specific region of the electromagnetic spectrum, with transitions ending at higher m values typically lying in the infrared region.

Transcript

00:01 And this problem, we're asked to consider two consecutive energy drops in hydrogen from an outer shell to the fourth shell.

00:08 So we're going to be considering a five to six drop, or excuse me, a five to four drop, and a six to four drop.

00:18 We're going to be using the balmer equation, which is one over lambda, equals the ryder constant, which is given.

00:27 One over m squared is the greater number, the smaller number, i should say, and the larger number.

00:37 I usually go one over lower energy level, higher energy level.

00:42 Where r is equal to 1 .097 times 10 to the minus 2, nanometers to the minus 1.

00:54 Okay.

00:56 We're going to call the wavelength, calculate the wavelengths in nanometers, and energy in kilojoules, i'm assuming, per mole, for the first two lines.

01:11 We'll be the 5 to 4 and the 6 to 4 drop.

01:15 Let's do the 5 to 4 drop first.

01:16 So this is pretty simple.

01:21 1 over lambda equals 1 .097 times 10 to the minus 2 nanometers to minus 1 times 1 over 4 squared minus 1 over 5 squared.

01:44 Okay.

01:46 So we're going to get a 1 over lambda that's equals to 2 .468 times 10 to the minus 4 nanometers to the minus 1.

02:01 And we'll get a lambda of 4 ,051 nanometers.

02:09 There's our wavelength for the first one.

02:12 Then let's do our energy, which will be planck's constant times c over lambda.

02:20 So it'll be 6 .626, just 10 to the minus 34th, dual seconds, times speed of light over lambda.

02:38 And remember, we've got to convert our nanometers to meters, so i'm going to take this times 10 to the minus 9...

Please give Ace some feedback