Nuclear Physics And Radioactivity Question 109
Question: Let $ v_{1} $ be the frequency of series limit of Lyman series, $ v_{2} $ the frequency of the first line of Lyman series and $ v_{3} $ the frequency of series limit of Balmer series. Then which of the following is correct?
Options:
A) $ v_{1}-v_{2}=v_{3} $
B) $ v_{2}-v_{1}=v_{3} $
C) $ v_{3}=\frac{1}{2}(v_{1}+v_{2}) $
D) $ v_{2}+v_{1}=v_{3} $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] Series limit means, the shortest possible wavelength (maximum photon energy) and first line means the largest possible wavelength (minimum photon energy) in the series $ v=C[ \frac{1}{n^{2}}-\frac{1}{m^{2}} ] $ (Where C is a constant) For series limit of Lyman series: $ n=1,m=\infty , $
$ \therefore v_{1}=C $ For first line of Lyman series: $ n=1,m=2, $
$ \therefore v_{2}=3C/4 $ For series limit of Balmer series: $ n=2, $ $ m=\infty $ ,
$ \therefore v_{3}=C/4 $
BETA
