Calculate the wave number for the longes | Class 11 Chemistry Chapter Structure of Atom, Structure of Atom NCERT Solutions

Welcome to the NCERT Solutions for Class 11 Chemistry - Chapter Structure of Atom. This page offers a step-by-step solution to the specific question from Exercise 1, Question 17: . With detailed answers and explanations for each chapter, students can strengthen their understanding and prepare confidently for exams. Ideal for CBSE and other board students, this resource will simplify your study experience.

Question 17:

Calculate the wave number for the longest wavelength transition in the Balmer series of atomic hydrogen.

Answer:

According to Balmer formula

ṽ=1/λ = RH[1/n12-1/n22]

For the Balmer series, ni = 2.

Thus, the expression of wavenumber(ṽ) is given by,

Wave number (ṽ) is inversely proportional to wavelength of transition. Hence, for the longest wavelength transition, ṽ has to be the smallest.

For ṽ to be minimum, nf should be minimum. For the Balmer series, a transition from ni = 2 to nf = 3 is allowed. Hence, taking nf = 3,we get:

ṽ= 1.5236 × 106 m–1


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Comments

  • nani
  • Oct 20, 2020

great


  • Musharraf Ahmed
  • Dec 08, 2019

Thanks


  • Musharraf Ahmed
  • Dec 08, 2019

Thanks


  • Unaiza
  • Oct 02, 2019

Thnku it hepled a lot..


  • Ck sahu
  • Sep 13, 2019

This is correct ans


  • Raghav Rai
  • Aug 27, 2019

Wait then value is 1.097.isit right


  • Misti
  • Aug 13, 2019

Very specific and precise answer .


  • Misti
  • Aug 13, 2019

Thnx a lot...it was best last minute tip I ever recieved


  • K surya
  • Jul 28, 2019

Change the background the answer was quite helpful


  • Dhriti
  • Jul 07, 2019

salute to this question.


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Welcome to the NCERT Solutions for Class 11 Chemistry - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 17: Calculate the wave number for the longest wavelength transition in the Balmer series of atomic hydro....