Explanation Given Info : The value of wavelength is 656 nm . Formula to calculate the wavelength of any spectral series is, 1 λ = R ( 1 n 1 2 − 1 n 2 2 ) (1) Here, λ is wavelength of spectral series. R is Rydberg’s constant. n 1 and n 2 are integer values for spectral series, which satisfies the condition of n 1 > n 2 . The value of Rydberg’s constant is 1.097 × 10 7 m − 1 . The given wavelength lies in the visible region and only Balmer series has visible radiations. Therefore, it is a Balmer series. Hence, the value of n f = 2 . Conversion of nm into m is done as, 1 nm = 10 − 9 m Therefore, conversion of 656 nm into m is done as, 1 nm = 10 − 9 m 656 nm = 656 × 10 − 9 m Substitute 1.097 × 10 7 m − 1 for R , 2 for n f and 656 × 10 − 9 for λ to find n i

5th Edition

ISBN: 9780134051802

Author: Walker

Publisher: PEARSON

See similar textbooks

Concept explainers

Quantum mechanics and hydrogen atom

Consider an electron of mass m moves with the velocity v in a hydrogen atom. If an electron is at a distance r from the proton, then the potential energy function of the electron can be written as follows:

Isotopes of Hydrogen Atoms

To understand isotopes, it's easiest to learn the simplest system. Hydrogen, the most basic nucleus, has received a great deal of attention. Several of the results seen in more complex nuclei can be seen in hydrogen isotopes. An isotope is a nucleus of t…

Mass of Hydrogen Atom

Hydrogen is one of the most fundamental elements on Earth which is colorless, odorless, and a flammable chemical substance. The representation of hydrogen in the periodic table is H. It is mostly found as a diatomic molecule as water H2O on earth. It is …

Question

Chapter 31, Problem 23PCE

To determine

The initial and final states when electron in hydrogen emits a photon with a wavelength of 656 nm .

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 31, Problem 22PCE

Chapter 31, Problem 24PCE

Students have asked these similar questions

3. By using the fact that around any closed loop the sum of the EMFS = the sum of the PDs. Write equations for the two loops shown in the cct below. 40 ΔΩ I₂ 4V (loop1 20 (loop2) 2v I+12 Use these equations to show that the current flowing through the 20 resistor is 0.75A

5. A potential divider circuit is made by stretching a 1 m long wire with a resistance of 0.1 per cm from A to B as shown. 8V A 100cm B sliding contact 5Ω A varying PD is achieved across the 5 Q resistor by moving the slider along the resistance wire. Calculate the distance from A when the PD across the 5 Q resistor is 6 V.

4. A voltmeter with resistance 10 kQ is used to measure the pd across the 1 kQ resistor in the circuit below. 6V 5ΚΩ 1ΚΩ V Calculate the percentage difference between the value with and without the voltmeter.

Chapter 31 Solutions

Physics (5th Edition)

Ch. 31.1 - Prob. 1EYU Ch. 31.2 - Prob. 2EYU Ch. 31.3 - Prob. 3EYU Ch. 31.4 - Prob. 4EYU Ch. 31.5 - Prob. 5EYU Ch. 31.6 - Prob. 6EYU Ch. 31.7 - Prob. 7EYU Ch. 31 - Prob. 1CQ Ch. 31 - Prob. 2CQ Ch. 31 - Prob. 3CQ

Ch. 31 - Prob. 4CQ Ch. 31 - Prob. 5CQ Ch. 31 - Prob. 6CQ Ch. 31 - Prob. 7CQ Ch. 31 - Prob. 8CQ Ch. 31 - Prob. 9CQ Ch. 31 - Prob. 1PCE Ch. 31 - Prob. 2PCE Ch. 31 - Prob. 3PCE Ch. 31 - Prob. 4PCE Ch. 31 - Prob. 5PCE Ch. 31 - Prob. 6PCE Ch. 31 - Prob. 7PCE Ch. 31 - Prob. 8PCE Ch. 31 - Prob. 9PCE Ch. 31 - Prob. 10PCE Ch. 31 - Prob. 11PCE Ch. 31 - Prob. 12PCE Ch. 31 - Prob. 13PCE Ch. 31 - Prob. 14PCE Ch. 31 - Prob. 15PCE Ch. 31 - Prob. 16PCE Ch. 31 - Prob. 17PCE Ch. 31 - Prob. 18PCE Ch. 31 - Prob. 19PCE Ch. 31 - Prob. 20PCE Ch. 31 - Prob. 21PCE Ch. 31 - Prob. 22PCE Ch. 31 - Prob. 23PCE Ch. 31 - Prob. 24PCE Ch. 31 - Prob. 25PCE Ch. 31 - Prob. 26PCE Ch. 31 - Prob. 27PCE Ch. 31 - Prob. 28PCE Ch. 31 - Prob. 29PCE Ch. 31 - Prob. 30PCE Ch. 31 - Prob. 31PCE Ch. 31 - Prob. 32PCE Ch. 31 - Prob. 33PCE Ch. 31 - Prob. 34PCE Ch. 31 - Prob. 35PCE Ch. 31 - Prob. 36PCE Ch. 31 - Prob. 37PCE Ch. 31 - Prob. 38PCE Ch. 31 - Prob. 39PCE Ch. 31 - Prob. 40PCE Ch. 31 - Prob. 41PCE Ch. 31 - Prob. 42PCE Ch. 31 - Prob. 43PCE Ch. 31 - Prob. 44PCE Ch. 31 - Prob. 45PCE Ch. 31 - Prob. 46PCE Ch. 31 - Prob. 47PCE Ch. 31 - Prob. 48PCE Ch. 31 - Prob. 49PCE Ch. 31 - Prob. 50PCE Ch. 31 - Prob. 51PCE Ch. 31 - Prob. 52PCE Ch. 31 - Give the electronic configuration for the ground...Ch. 31 - Prob. 54PCE Ch. 31 - Prob. 55PCE Ch. 31 - Prob. 56PCE Ch. 31 - The configuration of the outer electrons in Ni is...Ch. 31 - Prob. 58PCE Ch. 31 - Prob. 59PCE Ch. 31 - Prob. 60PCE Ch. 31 - Prob. 61PCE Ch. 31 - Prob. 62PCE Ch. 31 - Prob. 63PCE Ch. 31 - Prob. 64PCE Ch. 31 - Prob. 65PCE Ch. 31 - Prob. 66PCE Ch. 31 - Prob. 67PCE Ch. 31 - Prob. 68GP Ch. 31 - Prob. 69GP Ch. 31 - Prob. 70GP Ch. 31 - Prob. 71GP Ch. 31 - Prob. 72GP Ch. 31 - Prob. 73GP Ch. 31 - Prob. 74GP Ch. 31 - Prob. 75GP Ch. 31 - Prob. 76GP Ch. 31 - Prob. 77GP Ch. 31 - Prob. 78GP Ch. 31 - Prob. 79GP Ch. 31 - Prob. 80GP Ch. 31 - Prob. 81GP Ch. 31 - Prob. 82GP Ch. 31 - Prob. 83GP Ch. 31 - Prob. 84PP Ch. 31 - Prob. 85PP Ch. 31 - Prob. 86PP Ch. 31 - Prob. 87PP Ch. 31 - Prob. 88PP Ch. 31 - Prob. 89PP

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.

Similar questions

SEE MORE QUESTIONS

Recommended textbooks for you

College Physics
Physics
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Cengage Learning
College Physics
Physics
ISBN:9781285737027
Author:Raymond A. Serway, Chris Vuille
Publisher:Cengage Learning
Principles of Physics: A Calculus-Based Text
Physics
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Modern Physics
Physics
ISBN:9781111794378
Author:Raymond A. Serway, Clement J. Moses, Curt A. Moyer
Publisher:Cengage Learning
Inquiry into Physics
Physics
ISBN:9781337515863
Author:Ostdiek
Publisher:Cengage
Glencoe Physics: Principles and Problems, Student...
Physics
ISBN:9780078807213
Author:Paul W. Zitzewitz
Publisher:Glencoe/McGraw-Hill