Question

If the wavelength of H${}_{alpha}$ line of Balmer series is X A${}^o$, then the wavelenght of H${}_{beta}$ line of Balmer series is:

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

Answer 1

Answer: (B)

$X\frac{80}{108}$ A${}^o$

For Balmer series, $\frac{1}{\lambda }=R[\frac{1}{2^2}-\frac{1}{n_2^2}]$
For alpha line, $\frac{1}{{\lambda }_{\alpha }}=R[\frac{1}{2^2}-\frac{1}{3^2}]=\frac{1}{X}$
$\frac{1}{{\lambda }_{\alpha }}=R\times \frac{5}{36}=\frac{1}{X}$
$R=\frac{36}{5}\times \frac{1}{X}$...... (1)
For beta line, $\frac{1}{{\lambda }_{\beta }}=R[\frac{1}{2^2}-\frac{1}{4^2}]=\frac{3}{16}R$......(2)
Substitute equation (1) into (2) .
$\frac{1}{{\lambda }_{\beta }}=\frac{3}{16}R=\frac{3}{16}\times \frac{36}{5}\times \frac{1}{X}=\frac{108}{80}\times \frac{1}{X}$
${\lambda }_{\beta }=X\frac{80}{108}{A}^o$.