Question

The wavelength of the first member of Balmer series in hydrogen spectrum is $\lambda$. Calculate the wavelength of the first member of Lyman series in the same spectrum.

• A. $\left(5/27\right)\lambda$
• B. $\left(4/27\right)\lambda$
• C. $\left(27/5\right)\lambda$
• D. $\left(27/4\right)\lambda$

Answer 1

The correct option is A $\left(5/27\right)\lambda$

Given the wavelength of $1^{st}$ member of Balmer series in Hydrogen spectrum is ${}^{\prime }{\lambda }^{\prime }$

Rydberg's equation :-

$\overline{\upsilon }\frac{1}{\lambda }{R}_H^2\left[\frac{1}{n_1^2-\frac{1}{n_2^2}}\right]$

For Hydrogen $z=1$, for $1^{st}$ member of Balmer series $n_1=2,n_2=3$

$\Rightarrow \frac{1}{\lambda }=R_{H\times 1}[\frac{1}{2^2}-\frac{1}{3^2}]$

$\Rightarrow \frac{1}{\lambda }=R_H[\frac{1}{4}-\frac{1}{9}]$

$\Rightarrow \frac{1}{\lambda }=R_H\left(\frac{5}{36}\right)$ ------ $\left(1\right)$

For $1^{st}$ member of Lyman series (in same spectrum)

$n_1=1,n_2=2$

$\Rightarrow \frac{1}{{\lambda }^{\prime }}=R_{H\times 1}[\frac{1}{1^2}-\frac{1}{2^2}]$

$\Rightarrow \frac{1}{{\lambda }^{\prime }}=R_H[1-\frac{1}{4}]=R_H\left(\frac{3}{4}\right)$ ------ $\left(2\right)$

$\frac{\left(1\right)}{\left(2\right)}\Rightarrow \frac{\left(\frac{1}{\lambda }\right)}{\left(\frac{1}{{\lambda }^1}\right)}=\frac{R_H\left(\frac{5}{36}\right)}{R_H\left(\frac{3}{4}\right)}$

$\Rightarrow \frac{{\lambda }^1}{\lambda }=\frac{5}{36}\times \frac{4}{3}$

$\Rightarrow {\lambda }^1=\frac{5\lambda }{27}$

$\therefore {\lambda }^1=\frac{5\lambda }{27}$