Question

The ratio of the largest to shortest wavelengths in balmer series of hydrogen spectra is

Answer 1

when an hydrogen electron from level 6-level 2 it shows the transition.

where level ( 6) is higher energy level and (2) is lower energy level the wavelength of the radiation of any electron can be calculates by Rydberg equation. According to this equation as:

1/ w = R 1/ 4  -  1/ 36 where both 1/4 and 1/36 are reciprocal of square of lower and higher energy levels 2 and 6 in Balmer series and R" is Rydberg constant.

so value of both 1/4 and 1/36 is equal to 0.22 and value of R" is equal to

243,728 per meter.

by taking this value we can calculates the wavelength of any energy levels from 2  to 6 hope you will understand now

Answer 2

Shortest Transition of the Balmer series is from n₁ = 2 and n₂ = 3.

Longest Transition of the Balmer Series is from n₁ = 2 and n₂ = Infinity.

Shortest Transition means the longest wavelength and the Longest Transition means the shortest wavelength.

Using the Rydberg Equation for Hydrogen Atom,

1/λ₁ = R(1/n₁² - 1/n₂²)

1/λ₁ = R(1/4 - 1/9)

1/λ₁ = R(5/36)

λ₁ = 36/5R

Also,  1/λ₂ = R(1/n₁² - 1/n₂²)

1/λ₂ = R(1/4 - 1/Infinity²)

1/λ₂ = R/4

λ₂ = 4/R

Now, λ₁/λ₂ = 36/5R ÷ 4/R

 λ₁/λ₂ = 9/5

λ₁ : λ₂ = 9 : 5

Hope it helps.