Nuclear Physics And Radioactivity Question 30
Question: Hydrogen (H), deuterium , singly ionized helium $ (H{{e}^{+}}) $ and doubly ionized lithium $ (Li) $ all have one electron around the nucleus. Consider n =2 to n = 1 transition. The wavelengths of emitted radiations are $ {\lambda_{1}},\ {\lambda_{2}},\ {\lambda_{3}} $ and $ {\lambda_{4}} $ respectively. Then approximately [KCET 1994]
Options:
A) $ {\lambda_{1}}={\lambda_{2}}=4{\lambda_{3}}=9{\lambda_{4}} $
B) $ 4{\lambda_{1}}=2{\lambda_{2}}=2{\lambda_{3}}={\lambda_{4}} $
C) $ {\lambda_{1}}=2{\lambda_{2}}=2\sqrt{2}{\lambda_{3}}=3\sqrt{2}{\lambda_{4}} $
D) $ {\lambda_{1}}={\lambda_{2}}=2{\lambda_{3}}=3\sqrt{2}{\lambda_{4}} $
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Answer:
Correct Answer: A
Solution:
Using $ \Delta E\propto Z^{2} $ (∵ $ n_{1} $ and $ n_{2} $ are same)
Þ $ \frac{hc}{\lambda }\propto Z^{2} $
Þ $ \lambda Z^{2}=constant $
Þ $ {\lambda_{1}}Z_{1}^{2}={\lambda_{2}}Z_{2}^{2}={\lambda_{3}}Z_{3}^{2}={\lambda_{4}}Z_{^{4}}^{2} $
Þ $ {\lambda_{1}}\times 1={\lambda_{2}}\times 1^{2}={\lambda_{3}}\times 2^{2}={\lambda_{4}}\times 3^{3} $
Þ $ {\lambda_{1}}={\lambda_{2}}=4{\lambda_{3}}=9{\lambda_{4}} $ .
BETA
