Question

Let ${\nu }_1$ be the frequency of the series limit of the Lyman series, ${\nu }_2$ be the frequency of the first line of the Lyman series, and ${\nu }_3$ be the frequency of the series limit of the Balmer series, then

• A. ${\nu }_1-{\nu }_2={\nu }_3$
• B. ${\nu }_2-{\nu }_1={\nu }_3$
• C. ${\nu }_1+{\nu }_2=2{\nu }_3$
• D. ${\nu }_1+{\nu }_2={\nu }_3$

Answer 1

The correct option is A ${\nu }_1-{\nu }_2={\nu }_3$

$h{\nu }_1=13.6[\frac{1}{1}-\frac{1}{\infty }]$
${\nu }_1=C_0(whereC0=\frac{13.6}{h})$
Then
$h{\nu }_2=13.6[\frac{1}{1}-\frac{1}{2}]$
$\Rightarrow {\nu }_2=\frac{{\nu }_1}{2}$
Also $h{\nu }_3=13.6$
${\nu }_3=\frac{{\nu }_1}{2}$
$\Rightarrow {\nu }_2+{\nu }_3={\nu }_1$