Calculate the wavelength for the emissio | 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 56: . 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 56:

Calculate the wavelength for the emission transition if it starts from the orbit having radius 1.3225 nm and ends at 211.6 pm. Name the series to which this transition belongs and the region of the spectrum.

Answer:

The radius of the nth orbit of hydrogen-like particles = 0.529n2/Z Å

Now r1 = 1.3225 nm or 1322.5 pm = 52.9n12

And

r2 = 211.6pm = 52.9n22/Z

Therefore r1/r2 = 1322.5 / 211.6 = n12/n22

or n12/n22 = 6.25

or n1/n2 = 2.5

therefore n2 = 2 , n1 = 5.

Thus the transition is from 5th orbit to 2nd orbit. It belongs Balmer series

Wave number  for the transition is given by,

1.097 × 107 m–1 (1/22-1/52)

=1.097 x 107m-1 (21/100)

= 2.303 × 106 m–1

Wavelength (λ) associated with the emission transition is given by,

= 0.434 ×10–6 m

λ = 434 nm

the region is visible region


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Comments

  • Shinchan
  • Sep 09, 2017

Answers are very helpfull but pleas do give examples along with examples


Add Comment

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 56: Calculate the wavelength for the emission transition if it starts from the orbit having radius 1.322....