Atoms And Nuclei Question 73
Question: The first line of Balmer series has wavelength 6563 Å. What will be the wavelength of the first member of Lyman series [RPMT 1996]
Options:
A) 1215.4 Å
B) 2500 Å
C) 7500 Å
D) 600 Å
Show Answer
Answer:
Correct Answer: A
Solution:
$ \frac{1}{{\lambda_{Balmer}}}=R[ \frac{1}{2^{2}}-\frac{1}{3^{2}} ]=\frac{5R}{36} $ , $ \frac{1}{{\lambda_{Lyman}}}=R[ \frac{1}{1^{2}}-\frac{1}{2^{2}} ]=\frac{3R}{4} $
$ \therefore {\lambda_{Lyman}}={\lambda_{Balmer}}\times \frac{5}{27}=1215.4\ {\AA} $
BETA
