Question

If the shortest wavelength of the Lyman series of H atom is x, then the wavelength of the first line of Balmer series of H atom will be:

• A. $\frac{9x}{5}$
• B. $\frac{36x}{5}$
• C. $\frac{5x}{9}$
• D. $\frac{5x}{36}$

Answer 1

The correct option is B $\frac{36x}{5}$

Given: shortest wavelength of the Lyman series of H-atom is x.
For the shortest wavelength of Lyman series,
$n_1=1$ and $n_2=\infty$
$\therefore$ $\frac{1}{x}$$=$$R$[$\frac{1}{1^2}$$-$$\frac{1}{{\infty }^2}$]
So, $x$$=$$\frac{1}{R}$ or $R$$=$$\frac{1}{x}$

For the first line of Balmer series,
$n_1=2$ $n_2=3$
Wavelength of this line ($\lambda$) is given by:
$\frac{1}{\lambda }$$=$$R$($1^2$)($\frac{1}{2^2}$$-$$\frac{1}{3^2})\to eq.(1)$

Substituting , $R$$=$$\frac{1}{x}$ in equation (1), we get;

$\frac{1}{\lambda }$$=$$\frac{1}{x}$$\times$$\frac{5}{36}$
So, $\lambda$$=$$\frac{36x}{5}$