# The wavelength of the second line of Balmer series in the hydrogen spectrum is 4861 Å…

Q: The wavelength of the second line of Balmer series in the hydrogen spectrum is 4861 Å. The wavelength of the first line is

(a) $\displaystyle \frac{27}{20}\times 4861 A^o$

(b) $\displaystyle \frac{20}{27}\times 4861 A^o$

(c) $\displaystyle 20 \times 4861 A^o$

(d) $\displaystyle 4861 A^o$

Ans: (a)

Sol:

$\displaystyle \frac{1}{\lambda_2} = R (\frac{1}{2^2}-\frac{1}{4^2})$

$\displaystyle \frac{1}{\lambda_2} = R (\frac{1}{4}-\frac{1}{16})$

$\displaystyle \frac{1}{\lambda_2} = R (\frac{3}{16})$ …(i)

$\displaystyle \frac{1}{\lambda_1} = R (\frac{1}{2^2}-\frac{1}{3^2})$

$\displaystyle \frac{1}{\lambda_2} = R (\frac{1}{4}-\frac{1}{9})$

$\displaystyle \frac{1}{\lambda_2} = R (\frac{5}{36})$ …(ii)

On dividing (i) by (ii)

$\displaystyle \frac{\lambda_1}{\lambda_2} = \frac{3R/16}{5R/36}$

$\displaystyle \frac{\lambda_1}{\lambda_2} = \frac{27}{20}$

$\displaystyle \lambda_1 = \frac{27}{20}\times \lambda_2$

$\displaystyle \lambda_1 = \frac{27}{20}\times 4861 A^o$