Nuclear Physics And Radioactivity Question 93
Question: Hydrogen (H), deuterium (4), singly ionized helium ( $ H{{e}^{+}} $ and doubly ionized lithium (Li) all have one electroi 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
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:
[a] Using $ \Delta E\propto Z^{2} $ (
$ \therefore $ $ n_{1} $ and $ n_{2} $ are same)
$ \Rightarrow \frac{hc}{\lambda }\propto Z^{2}\Rightarrow \lambda Z^{2}=constant $
$ \Rightarrow {\lambda_1}Z^{2}_1$=${\lambda_2}Z^{2}_2$=${\lambda_3}Z^{2}_3$=${\lambda_4}Z^{2}_4 $
$ \Rightarrow {\lambda_{1}}\times 1={\lambda_{2}}\times 1^{2}={\lambda_{3}}\times 2^{2}={\lambda_{4}}\times 3^{3} $
$ \Rightarrow {\lambda_{1}}={\lambda_{2}}=4{\lambda_{3}}=9{\lambda_{4}} $
BETA
