Question

The relation between ${\lambda }_3:$ wavelength of series limit of Lyman series, ${\lambda }_2:$ the wavelength of the series limit of Balmer series & ${\lambda }_1:$ the wavelength of first line of Lyman series is

• A. ${\lambda }_1={\lambda }_2+{\lambda }_3$
• B. ${\lambda }_3={\lambda }_1+{\lambda }_2$
• C. ${\lambda }_2={\lambda }_3-{\lambda }_1$
• D. $\frac{1}{{\lambda }_1}-\frac{1}{{\lambda }_2}=\frac{1}{{\lambda }_3}$

Answer 1

The correct option is D $\frac{1}{{\lambda }_1}-\frac{1}{{\lambda }_2}=\frac{1}{{\lambda }_3}$

$\begin{array}{cc}{\lambda }_1=& \text{wavelength of series limit of Lyman}\\ {\lambda }_2=& \text{wavelength of series limit of balmer}\\ {\lambda }_3& =\text{warelength of first line of lyman}\end{array}$

$\begin{array}{c}\text{for Lyman, series limit}\\ \begin{array}{cc}& n=1\text{to}{n}_2=\infty \\ \frac{1}{{\lambda }_1}& =R{z}^2(\frac{1}{1}-\frac{1}{\infty })\\ \frac{1}{{\lambda }_1}& =R{z}^2-\left(i\right)\end{array}\end{array}$

$\begin{array}{c}\text{for balmer, series limit}\\ \begin{array}{c}{n}_1=2,\text{to}{n}_2=\infty \\ \frac{1}{{\lambda }_2}=R{z}^2(\frac{1}{4}-\frac{1}{\infty })\\ \frac{1}{{\lambda }_2}=\frac{R{z}^2}{4}-\left(ii\right)\end{array}\end{array}$

$\begin{array}{c}\text{for First line of Lyman}\\ \begin{array}{cc}{n}_1& =1n_2=2\\ \frac{1}{{\lambda }_3}& =R{z}^2(\frac{1}{1}-\frac{1}{4})\\ & \frac{1}{{\lambda }_3}=R\cdot z^2\frac{3}{4}-\left(iii\right)\\ \text{from}\left(i\right),\left(ii\right)& \text{and}\left(iii\right)\end{array}\end{array}$

$\frac{1}{{\lambda }_1}-\frac{1}{{\lambda }_2}=\frac{1}{{\lambda }_3}$