Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 18: Problem 2539

In terms of Rydberg constant \(R\). The wave number of first Balmer line is (A) \((5 \mathrm{R} / 36)\) (B) \((8 \mathrm{R} / 9)\) (C) \(\mathrm{R}\) (D) \((8 \mathrm{R} / 20)\)

Short Answer

Expert verified
The wave number of the first Balmer line in terms of the Rydberg constant R can be calculated using the Rydberg formula for wave number \(\bar\nu = R \times ( \frac{1}{n_1^2} - \frac{1}{n_2^2})\). For the first Balmer line, \(n_1 = 2\) and \(n_2 = 3\). Plugging these values and simplifying the expression, we get \(\bar\nu = \frac{5R}{36}\), which corresponds to option (A).

Step by step solution

01

Recall the Rydberg formula for wave number

The Rydberg formula for wave number is given by: \(\bar\nu = R \times ( \frac{1}{n_1^2} - \frac{1}{n_2^2})\) Here, \(\bar\nu\) is the wave number, \(R\) is the Rydberg constant, \(n_1\) is the lower energy level, and \(n_2\) is the higher energy level.
02

Apply the formula for the first Balmer line

For the first Balmer line, we have \(n_1 = 2\) (as Balmer series transitions end at the second energy level) and \(n_2 = 3\) (as this is the initial energy level for the first Balmer line). Plugging these values into the Rydberg formula, we get: \(\bar\nu = R \times ( \frac{1}{2^2} - \frac{1}{3^2})\)
03

Simplify the expression

Simplify the expression obtained in step 2: \(\bar\nu = R \times ( \frac{1}{4} - \frac{1}{9})\) To find a common denominator for proceeding with the subtraction, we can use 36 as the denominator for both (as it is the least common multiple of 4 and 9) fractions: \(\bar\nu = R \times ( \frac{9}{36} - \frac{4}{36})\) Continue with the subtraction: \(\bar\nu = R \times \frac{9-4}{36}\) \(\bar\nu = R \times \frac{5}{36}\) Thus, the wave number of the first Balmer line in terms of the Rydberg constant R is: \(\boxed{\bar\nu = \frac{5R}{36}}\) This corresponds to option (A).

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

The half time of a radioactive substance is \(20 \mathrm{~min}\), difference between Points of time when it is \(33 \%\) disintegrated and \(67 \%\) disintegrated is approximately (A) \(10 \mathrm{~min}\) (B) \(20 \mathrm{~min}\) (C) \(40 \mathrm{~min}\) (D) \(30 \mathrm{~min}\)

\(\mathrm{A}\) and \(\mathrm{B}\) are two radioactive substance whose half lives are 1 and 2 years respectively. Initially \(10 \mathrm{~g}\) of \(\mathrm{A}\) and \(1 \mathrm{~g}\) of \(\mathrm{B}\) is taken. The time after which they will have same quantity remaining is (A) \(3.6\) years (B) 7 years (C) \(6.6\) years (D) 5 years

The binding energy Per nucleon of deuteron $\left({ }^{2}{ }_{1} \mathrm{H}\right)\( and Lielium nucleus \){ }_{2}{ }^{4}{ }_{2} \mathrm{He}$ ) is \(1.1 \mathrm{MeV}\) and \(7.0 \mathrm{MeV}\). respectively. If two deuteron react to form a single helium nucleus, the energy released is (A) \(23.6 \mathrm{MeV}\) (B) \(26.9 \mathrm{MeV}\) (C) \(13.9 \mathrm{MeV}\) (D) \(19.2 \mathrm{MeV}\)

The shape of the graph \(\ln 1 \rightarrow t\) is (A) straight Line (B) Parabolic curve (C) Hyperbole curve (D) random shape curve

In Bohr model the hydrogen atom, the lowest orbit corresponds to (A) Infinite energy (B) zero energy (C) The minimum energy (D) The maximum energy
See all solutions

Recommended explanations on English Textbooks

Argumentative Essay

Read Explanation

English Language Study

Read Explanation

Linguistic Terms

Read Explanation

5 Paragraph Essay

Read Explanation

Creative Story

Read Explanation

Language and Social Groups

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