Question

ratio of wavelength of first line of Balmer series to first line of lyman series is

Answer 1

ur answer should be 27/5
do u need explaination ??

Answer 2

Answer:  Ratio of minimum wavelength of lyman and balmer series will be 27: 5.

Explanation:  

$E=\frac{hc}{\lambda}$

$\lambda$ = Wavelength of radiation

E= energy

1. The wavelength of first line of balmer series i.e the electron will jump from n=2 to n=3.

Using Rydberg's Equation:

$\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )$

Where,

$\lambda$ = Wavelength of radiation

$R_H$ = Rydberg's Constant

$n_f$ = Higher energy level = 3

$n_i$= Lower energy level = 2 (balmer series)

Putting the values, in above equation, we get

$\frac{1}{\lambda_{lyman}}=R_H\left(\frac{1}{2^2}-\frac{1}{3^2} \right )$

$\lambda_{lyman}=\frac{5}{36R_H}$

2. The wavelength of first line of lyman series i.e the electron will jump from n=1 to n=2

.$\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )$

Where,

$\lambda$ = Wavelength of radiation

$R_H$ = Rydberg's Constant

$n_f$ = Higher energy level = 2

$n_i$= Lower energy level = 1 (Lyman series)

Putting the values, in above equation, we get

$\frac{1}{\lambda_{balmer}}=R_H\left(\frac{1}{1^2}-\frac{1}{2^2} \right )$

$\lambda_{balmer}=\frac{3}{4R_H}$

Thus $\frac{\lambda_{balmer}}{\lambda_{lyman}}=\frac{\frac{3}{4R_H}}{\frac{5}{36R_H}}=\frac{27}{5}$