Atoms And Nuclei Question 131
Question: The shortest wavelength in the Lyman series of hydrogen spectrum is 912 Å corresponding to a photon energy of 13.6 eV. The shortest wavelength in the Balmer series is about [MP PMT 2004]
Options:
A) 3648 Å
B) 8208 Å
C) 1228 Å
D) 6566 Å
Show Answer
Answer:
Correct Answer: A
Solution:
In Lyman series $ {{({\lambda _{\min }})} _{L}}=\frac{1}{R} $ and $ {{({\lambda _{\min }})} _{B}}=\frac{1}{R}\left(1-\frac{1}{4}\right)^{-1} $ Þ $ {{({\lambda _{\min }})} _{B}}=4\times {{({\lambda _{\min }})} _{L}}=4\times 912=3648,{\AA} $
BETA
