LIVE Course for free

Rated by 1 million+ students
Get app now
0 votes
655 views
in Chemistry by (47.4k points)
closed by

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

1 Answer

+1 vote
by (45.8k points)
selected by
 
Best answer

Wave number \(\bar{V}\) = 109677 \(\big(\frac{1}{n_1^2}-\frac{1}{n_2^2}\big)\)

n1 = 2

n2 = 3 For Balmer series

\(\therefore\) \(\bar{V}\) = 109677 \(\Big(\frac{1}{(2)^2}-\frac{1}{(3)^2}\Big)\)

= 109677 \(\Big(\frac{1}{4}-\frac{1}{9}\Big)\)

= 109677 \(\Big(\frac{9-4}{36}\Big)\)

= 109677 x \(\frac{5}{36}\)

= 1.523 × 104 cm−1

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...