Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

If the series limit of wavelength of the Lyman series for the hydrogen atom is \(912 \wedge\) then the series limit of wavelength for the Balmer series of the hydrogen atom is (1) \(912 \AA\) (2) \(912 \times 2 \AA\) (3) \(912 \times 4 \AA\) (4) \(912 / 2 \AA\)

Short Answer

Expert verified
Option (3) 912 × 4 Å

Step by step solution

01

- Understand the Given Information

The problem provides the series limit of the Lyman series for the hydrogen atom as 912 Å.
02

- Recall the Series Formula

The Lyman series corresponds to electronic transitions where the electron falls to the n=1 level (ground state). The Balmer series corresponds to electronic transitions where the electron falls to the n=2 level.
03

- Use the Relationship Between Series Limits

The series limit for the Lyman series is given as 912 Å, which is the transition to the ground state (n=1). For the Balmer series, the transitions are to the n=2 level. The ratio of the series limits is given by the square of the ratio of the levels: \[\frac{1}{\text{Balmer limit}} = \frac{1}{\text{Lyman limit}} \times \frac{1}{4} \] Therefore, the wavelength of the Balmer series limit is 4 times that of the Lyman series limit.
04

- Calculate the Balmer Limit

Given the Lyman series limit is 912 Å, the Balmer series limit will be: \[\text{Balmer limit} = 912 \times 4 \text{ Å} = 3648 \text{ Å}\]
05

- Select the Correct Option

From the calculation, the series limit of the wavelength for the Balmer series of the hydrogen atom is 3648 Å. This matches the given option (3).

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.

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

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Lyman series
The Lyman series pertains to electronic transitions in a hydrogen atom that lead to the release of photons. This series involves electrons descending from higher energy levels (n >= 2) to the ground state (n=1).
When these electrons fall to the lowest energy level, they emit ultraviolet radiation, making the Lyman series significant in UV spectroscopy.
Key points to remember about the Lyman series:
  • Transitions end at the principal quantum number n=1.
  • Emits ultraviolet light.
  • Key wavelength limit is 912 ångströms (Å).
Understanding these aspects will aid in comprehending its role in hydrogen's spectral lines.
Balmer series
The Balmer series is associated with electronic transitions where electrons fall to the second energy level (n=2) in a hydrogen atom. Unlike the Lyman series, the Balmer series ejects photons in the visible part of the electromagnetic spectrum.
Key points to remember about the Balmer series:
  • Transitions end at the principal quantum number n=2.
  • Emits visible light.
  • Key wavelength components are visible in the range 3648 Å (for the series limit).
This series is highly significant in understanding the visible spectra of hydrogen and setting the foundation for detecting other elements in stars.
Electronic transitions
Electronic transitions describe movements of electrons between different energy levels within an atom. When an electron jumps from a higher energy level to a lower one, it emits a photon corresponding to the energy difference.
Key points to remember about electronic transitions:
  • Energy levels are quantized, meaning electrons can only exist at specific energy values.
  • The frequency and wavelength of the emitted photon directly correlate to the energy gap between the levels.
  • Each transition's characteristics define the spectral series (Lyman, Balmer, etc.).
Understanding these transitions is fundamental for interpreting atomic spectra and analyzing the emission and absorption lines in various elements.

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

Which of the following statements is false? (1) de Broglic wavelength associated with matter particle is inversely proportional to momentum. (2) de Broglic equation is a relationship between a moving particle and its momentum. (3) de Broglie equation suggests that an electrons has particle wave nature. (4) de Broglie equation is applicable to electrons only.

Considering the three electronic transitions \(n=2 \rightarrow\) \(n=1, n=3 \rightarrow n=2\) and \(n=4 \rightarrow n=3\) for the hydrogen at which one of the following is true. (1) The photon emitted in the transition \(n=4\) to \(n=3\) would have the longest wavelength. (2) The photon emitted in the transition \(n=2\) to \(n=1\) would have the longest wavelength. (3) The transition from \(n=3\) to \(n=1\) is forbidden. (4) The electron does not experience any change in orbit radius for any of these transitions.

Which of the following about the electron orbital is false? (1) No orbital can contain more than two electrons. (2) If two electrons occupy the same orbital, they must have different spins. (3) No two orbitals in an atom can have the same energy. (4) The number of orbitals in different subshells is not the same.

The binding energy of the electron in the lowest orbit of the hydrogen atom is \(13.6 \mathrm{cV}\). The energies required in \(\mathrm{cV}\) to remove an electron from three lowest orbits of the hydrogen atom arc (1) \(13.6,6.8,8.4 \mathrm{eV}\) (2) \(13.6,10.2,3.4 \mathrm{eV}\) (3) \(13.6,27.2,40.8 \mathrm{eV}\) (4) \(13.6,3.4,1.5 \mathrm{eV}\)

An electron that will have the highest energy in the set is (1) \(3,2,1, \frac{1}{2}\) (2) \(4,2,-1, \frac{1}{2}\) (3) \(4,1,0,-\frac{1}{2}\) (4) \(5,0,0, \frac{1}{2}\)

See all solutions

Recommended explanations on Chemistry Textbooks

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