Question

The wavelength of $H_{\alpha }$ line of Balmer series is $X\mathring{A}$. What is the $X$ of $H_{\beta }$ line of Balmer series

• A. $X\frac{108}{80}\mathring{A}$
• B. $X\frac{80}{108}\mathring{A}$
• C. $\frac{1}{X}\frac{80}{108}\mathring{A}$
• D. $\frac{1}{X}\frac{108}{80}\mathring{A}$

Answer 1

Answer: (B)

$H_{\alpha }$ line of Balmer series means first line of Balmer series
$n_1=2,n_2=3$
$\bar{v}=\frac{1}{{\lambda }_{\alpha }}=R\left(\frac{1}{2^2}-\frac{1}{3^2}\right)=\frac{5R}{36}$
$\therefore {\lambda }_{\alpha }=\frac{36}{5R}=X$
$H_{\beta }$ line of Balmer series means, second line of Balmer series, $n_1=2,n_2=4$
$\bar{v}=\frac{1}{{\lambda }_{\beta }}=R\left(\frac{1}{2^2}-\frac{1}{4^2}\right)=\frac{3R}{16}$
$\therefore {\lambda }_{\beta }=\frac{16}{3R}=X$
When $\frac{36}{3R}=X$
Then $\frac{16}{3R}=\frac{X\times 5R\times 36}{36\times 3R}=\frac{80X}{108}\mathring{A}$