Atoms And Nuclei Question 320
Question: The energy of electron in the nth orbit of hydrogen atom is expressed as $ E_{n}=-\frac{13.6}{n^{2}}eV. $ The shortest and longest wavelength of Lyman series will be
Options:
A) $ 910\overset{o}{\mathop{A}},,1213\overset{o}{\mathop{A}}, $
B) $ 5463\overset{o}{\mathop{A}},,7858\overset{o}{\mathop{A}}, $
C) $ 1315\overset{o}{\mathop{A}},,1530\overset{o}{\mathop{A}},~~ $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
- $ \frac{1}{{\lambda_{\max }}}=R[ \frac{1}{{{( 1 )}^{2}}}-\frac{1}{{{( 2 )}^{2}}} ]\Rightarrow {\lambda_{\max }}=\frac{4}{3R}\approx 1213\overset{o}{\mathop{A}}, $ and $ \frac{1}{{\lambda_{\min }}}=R[ \frac{1}{{{( 1 )}^{2}}}-\frac{1}{\infty } ]\Rightarrow {\lambda_{\min }}=\frac{1}{R}\approx 910\overset{o}{\mathop{A}}, $
