Calculate the wavelength for the emissio | Class 11 Chemistry Chapter Structure of Atom, Structure of Atom NCERT Solutions

Question:

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


Study Tips for Answering NCERT Questions:

NCERT questions are designed to test your understanding of the concepts and theories discussed in the chapter. Here are some tips to help you answer NCERT questions effectively:

  • Read the question carefully and focus on the core concept being asked.
  • Reference examples and data from the chapter when answering questions about Structure of Atom.
  • Review previous year question papers to get an idea of how such questions may be framed in exams.
  • Practice answering questions within the time limit to improve your speed and accuracy.
  • Discuss your answers with your teachers or peers to get feedback and improve your understanding.

Comments

  • Shinchan
  • 2017-09-09 20:41:40

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


Comment(s) on this Question

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....