Question

The wavelength of the first member of the Balmer series in hydrogen spectrum is $x$ $\mathring{A}$. Then the wavelength (in $\mathring{A}$) of the first member of Lyman series in the same spectrum is

• A. $\frac{5}{27}x$
• B. $\frac{4}{3}x$
• C. $\frac{27}{5}x$
• D. $\frac{5}{36}x$

Answer 1

Answer: (A)

$\frac{5}{27}x$

$\begin{array}{c}\text{Balmer series : transition take place from}\\ n=2\text{to}n=3,4,5,\text{so on}\\ \text{for first member of balmer}\\ n_1=2\text{to}{n}_2=3\end{array}$

$\lambda =x\dot{A}$

$\begin{array}{c}\frac{1}{\lambda }=R{z}^2(\frac{1}{n_{1}2}-\frac{1}{n_2^2})\\ \frac{1}{x}=R{z}^2(\frac{1}{4}-\frac{1}{9})\\ \frac{1}{x}=R{z}^2\left(\frac{5}{36}\right)-\left(1\right)\end{array}$

$\begin{array}{c}\text{For first member of Lyman}\\ \begin{array}{cc}{n}_1& =3n_2=2\\ \frac{1}{\lambda }& =R_2^2(\frac{1}{1}-\frac{1}{4})\\ \frac{1}{\lambda }& =R{z}^2\left(\frac{3}{4}\right)\end{array}\end{array}$

$\begin{array}{c}\text{divide (i)}÷\text{(ii)}\\ \begin{array}{c}\frac{\lambda }{x}=\left(\frac{5}{36}\right)\left(\frac{4}{3}\right)\\ \lambda =\left(\frac{5}{27}\right)x\end{array}\end{array}$