Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 6: Problem 77

In 1885 , Johann Balmer, a mathematician, derived the following relation for the wavelength of lines in the visible spectrum of hydrogen $$ \lambda=\frac{364.5 \mathrm{n}^{2}}{\left(\mathrm{n}^{2}-4\right)} $$ where \(\lambda\) is in nanometers and \(\mathbf{n}\) is an integer that can be \(3,4,5, \ldots\). Show that this relation follows from the Bohr equation and the equation using the Rydberg constant. Note that in the Balmer series, the electron is returning to the \(\mathbf{n}=2\) level.

Short Answer

Expert verified
Question: Show that Balmer's relation follows from the Bohr equation and the Rydberg constant. Answer: Balmer's relation, \(\lambda = \frac{364.5 n_i^2}{n_i^2 -4}\), is derived from the Bohr energy equation, \(E_n = -\frac{13.6 \mathrm{eV}}{n^2}\), and the Rydberg formula, \(\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)\). The derivation involves combining these equations through energy conservation and Planck's equation, resulting in a formula for the wavelength of emitted photons in the Balmer series.

Step by step solution

01

Write the Bohr's energy equation

The energy of an electron in any energy level \((n)\) in the hydrogen atom according to Bohr's theory is: $$ E_n = -\frac{13.6 \mathrm{eV}}{n^2} $$
02

Write the equation involving the Rydberg constant

In the case of hydrogen, the Rydberg formula for the wavelength \((\lambda)\) of emitted photons is given by: $$ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) $$ where \(R \approx 1.097 \times 10^7 \mathrm{m}^{-1}\) is the Rydberg constant, \(n_1\) and \(n_2\) are the initial and final energy levels, respectively (with \(n_2 > n_1\)).
03

Balmer series - return to n=2 level

In Balmer series, the electron returns to the \(n=2\) level. Let's denote the initial energy level as \(n_i\). Then \(n_1 = n_i\) and \(n_2=2\). Substitute these values in the equation from step 2: $$ \frac{1}{\lambda} = R \left( \frac{1}{n_i^2} - \frac{1}{2^2} \right) $$
04

Calculate the energy difference using the Bohr's energy equation

The energy difference between the two energy levels is the difference between the energies of the levels (\(\Delta E\)): $$ \Delta E = E_2 - E_{n_i} $$ Using Bohr's energy equation, we get: $$ \Delta E = -\frac{13.6 \mathrm{eV}}{2^2} - \left( -\frac{13.6 \mathrm{eV}}{n_i^2} \right) $$
05

Relate the energy difference to the wavelength using the Planck's equation

The energy difference between the two levels is equal to the energy of the emitted photon: $$ \Delta E = h \cdot \nu $$ where \(h\) is the Planck's constant and \(\nu\) is the frequency of the photon. We can rewrite the frequency as \(\nu = \frac{c}{\lambda}\), where \(c\) is the speed of light. $$ \Delta E = h\frac{c}{\lambda} $$
06

Combine the equations to find Balmer's relation

First, rewrite the equation from Step 5 in terms of \(\lambda\): $$ \lambda = \frac{hc}{\Delta E} $$ We have found \(\Delta E\) in Step 4. Substitute this expression into the above equation: $$ \lambda = \frac{hc}{ -\frac{13.6 \mathrm{eV}}{2^2} - \left( -\frac{13.6 \mathrm{eV}}{n_i^2} \right)} $$ Simplify this equation, and convert the Planck's constant times the speed of light from Joule meters to electron volt nanometers (1 eV = 1240 nm): $$ \lambda = \frac{364.5 n_i^2}{n_i^2 -4} $$ This is the Balmer's relation. We can see that it is derived from the Bohr theory and the equation involving the Rydberg constant.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Write the ground state electron configuration for (a) \(\mathrm{S}, \mathrm{S}^{2-}\) (b) \(\mathrm{Al}, \mathrm{Al}^{3+}\) (c) \(\mathrm{V}, \mathrm{V}^{4+}\) (d) \(\mathrm{Cu}^{+}, \mathrm{Cu}^{2+}\)

Criticize or comment on the following statements: (a) The energy of a photon is inversely proportional to its wavelength. (b) The energy of the hydrogen electron is inversely proportional to the quantum number \(\ell\). (c) Electrons start to enter the fifth principal level as soon as the fourth is full.

Explain the difference between (a) the Bohr model of the atom and the quantum mechanical model. (b) wavelength and frequency. (c) the geometries of the three different p orbitals.

Give the symbol of the main-group metals in period 4 with the following number of unpaired electrons per atom. (Transition metals are not included.) (a) 0 (b) 1 (c) 2 (d) 3

Give the number of unpaired electrons in an atom of (a) phosphorus (b) potassium (c) plutonium (Pu)
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Physical Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Chemical Analysis

Read Explanation

Chemical Reactions

Read Explanation

Nuclear Chemistry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free