JEE Main On 8 April 2017 Question 1
Question: If the shortest wavelength in Lyman series of hydrogen atom is A, then the longest wavelength in Paschen series of $ H{e^{+}} $ is : [JEE Online 08-04-2017]
Options:
A) $ \frac{5A}{9} $
B) $ \frac{36A}{7} $
C) $ \frac{36A}{5} $
D) $ \frac{9A}{5} $
Show Answer
Answer:
Correct Answer: B
Solution:
- Shortest wavelength is corresponding to best ine
$ \therefore $ $ n _{L}=1 $ (Lyman series) $ n _{H}=\infty $ (infinite) $ \frac{1}{A}=r\times {{(1)}^{2}}{ \frac{1}{12}-\frac{1}{2} }=R $ Longest wavelength $ \equiv $ 1st Line
$ \therefore $ = 3 $ n _{H}=4 $ $ \frac{1}{\lambda }=r\times {{(2)}^{2}}{ \frac{1}{3^{2}}-\frac{1}{4^{2}} }=\frac{r\times 7}{36} $ $ \lambda =\frac{36A}{7} $
