Atomic Structure Question 230
Question: The wavelength of $ {H_{\alpha }} $ line of Balmer series is X $ \overset{o}{\mathop{A}}, $ . What is the X of $ {H_{\beta }} $ line of Balmer series.
Options:
A) $ X\frac{108}{80}\overset{o}{\mathop{A}}, $
B) $ X\frac{80}{108}\overset{o}{\mathop{A}}, $
C) $ \frac{1}{X}\frac{80}{108}\overset{o}{\mathop{A}}, $
D) $ \frac{1}{X}\frac{108}{80}\overset{o}{\mathop{A}}, $
Show Answer
Answer:
Correct Answer: B
Solution:
- $ {H_{\alpha }} $ line of Balmer series means first line of Balmer series. $ n_1=2,n_2=3 $
$ \overline{v}=\frac{1}{{\lambda_{\alpha }}}=R( \frac{1}{2^{2}}-\frac{1}{3^{2}} )=\frac{5R}{36} $
$ \therefore {\lambda_{\alpha }}=\frac{36}{5R}=X $
$ {H_{\beta }} $ line of Balmer series means, second line of Balmer series, $ n_1=2,n_2=4 $
$ \overline{v}=\frac{1}{{\lambda_{\beta }}}=R( \frac{1}{2^{2}}-\frac{1}{4^{2}} )=\frac{3R}{16} $
$ \therefore {\lambda_{\beta }}=\frac{16}{3R}=X $ when $ \frac{36}{3R}=X $ then $ \frac{16}{3R}=\frac{X\times 5R\times 16}{36\times 3R}=\frac{80X}{108}\overset{{}^\circ }{\mathop{A}}, $
BETA
