Question

Approx wavelength of first line of paschen series of hydrogen atom

Answer 1

The first transition in the Paschen series corresponds to

ni=4 → nf=3

In this transition, the electron drops from the fourth energy level to the third energy level.

You will have

1λ1=R⋅(132−142)

The second transition in the Paschen series corresponds to

ni=5 → nf=3

This time, you have

1λ2=R⋅(132−152)

Now, to get the ratio of the first line to that of the second line, you need to divide the second equation by the first one.

1λ21λ1=R⋅(132−142)R⋅(132−152)

This will be equivalent to

λ1λ2=19−12519−116

λ1λ2=25−925⋅916−916⋅9=1625⋅9⋅16⋅97=16225⋅7

Therefore, you can say that you have

λ1λ2=256175