Atoms And Nuclei Question 89
Question: If $ {\lambda_{max}} $ is 6563 Å, then wave length of second line for Balmer series will be [RPMT 2000]
Options:
A) $ \lambda =\frac{16}{3R} $
B) $ \lambda =\frac{36}{5R} $
C) $ \lambda =\frac{4}{3R} $
D) None of the above
Show Answer
Answer:
Correct Answer: A
Solution:
For Balmer series $ \frac{1}{\lambda }=R( \frac{1}{2^{2}}-\frac{1}{n^{2}} ) $ where n = 3, 4, 5
For second line n = 4
So $ \frac{1}{\lambda }=R( \frac{1}{2^{2}}-\frac{1}{4^{2}} )=\frac{3}{16}R\Rightarrow \lambda =\frac{16}{3R} $
