Atoms And Nuclei Question 442
Question: The extreme wavelengths of Paschen series are
Options:
A) $ 0.365\mu m $ and $ 0.565\mu m $
B) $ 0.818\mu m $ and $ 1.89\mu m $
C) $ 1.45\mu m $ and $ 4.04\mu m $
D) $ 2.27\mu m $ and $ 7.43\mu m $
Show Answer
Answer:
Correct Answer: B
Solution:
(b) In Paschen series
$ \frac{1}{{\lambda_{\max }}}=R[ \frac{1}{{{(3)}^{2}}}-\frac{1}{{{(4)}^{2}}} ] $
Þ $ {\lambda_{\max }}=\frac{144}{7R}=\frac{144}{7\times 1.1\times 10^{7}}=1.89\times {{10}^{-6}}m=1.89,\mu m $
Similarly
$ {\lambda_{\min }}=\frac{9}{R}=\frac{9}{1.1\times 10^{7}}=0.818,\mu m $
