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

The ratio of the radius of the first Bohr orbit for the electron orbiting around the hydrogen nucleus that of the electron orbiting around the deuterium nucleus is approximately (1) \(1: 1\) (2) \(1: 2\) (3) \(2: 1\) (4) \(1: 4\)

Which of the following statement is false? (1) The orbit with more number of nodal planes will be of more energy. (2) The orbital with two angular nodes (nodal planes) is f-orbital. (3) The zero probability of finding the electron in \(\mathrm{p}_{x}-\) orbital is in \(y-z\) plane. (4) The orbital which do not has angular nodes is s-orbital.

A surface ejects electrons when hit by violet light but not when hit by yellow light. Will electrons be ejected if the surface is hit by red light? (1) Yes (2) No (3) Yes, if the red beam is quite intense (4) Yes, if the red beam continues to fall upon the surface for some time

Among the following the correct statements are I. Neutrons are not deflected by any magnetic field. II. Cathode rays can be deflected by both magnetic and electric fields. III. Atomic weight of an element is not necessarily be a whole number because it is average isotopic weight. IV. The constancy of \(e / m\) ratio for electron shows that electron mass is \(1 / 1837\)th of proton. (1) I, II, III and IV (2) 1,11 and III (3) 1\(], 111\) and IV (4) \(1,11 \mid\) and IV

Which of the following statements is wrong? (1) In the hydrogen spectrum least energetic series is Pfund series. (2) The lines of longest wavelength in the Balmer series correspond to the transition between \(n=3\) and \(n=2\) levels. (3) The dark lines in a spectrum are produced by existing gases to very high energy levels. (4) The wave number of infinity line in Lyman series of hydrogen spectrum is \(9 \mathrm{R} / 3\).

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