Question

the ratio of minimum wavelength of lyman and balmer series will be:-

Answer 1

Put n2=infinity in Rydbergs formula and solve with

Lyman(n1=1)

Balmer(n1=2)

You will get 2 equations

Solve and get the answer:)

Just like this

I hope I got it right.

P.S. I’ve assumed you’re talking Hydrogen here.

Answer 2

Answer: Ratio of minimum wavelength of lyman and balmer series will be 1: 4.

Explanation:-

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

$\lambda$ = Wavelength of radiation

E= energy

For wavelength to be minimum, energy would be maximum, i.e the electron  will jump to infinite level.

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 = $\infty$

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

Putting the values, in above equation, we get

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

$\lambda_{lyman}=\frac{1}{R_H}$

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 = $\infty$

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

Putting the values, in above equation, we get

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

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

Thus $\frac{\lambda_{lyman}}{\lambda_{balmer}}=\frac{\frac{1}{R_H}}{\frac{4}{R_H}}=\frac{1}{4}$