The ratio of series limit wavlength of Balmer series to wavelength of first line of paschen series is ............

My try

I am getting the answer as (7/128) but the answer is (7/36)

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Best answer

You are using the wrong numbers for the Balmer Series.

The Rydbergy Formula for all of the spectral series in hydrogen-like atoms is

$\frac{1}{\lambda}=R Z^2 (\frac{1}{m^2}-\frac{1}{n^2})$.

For the Balmer Series $m=2$ and for the Paschen Series $m=3$. It is the Lyman Series which has $m=1$.

The series limit for the Balmer Series has $n \to \infty$ :

$\frac{1}{\lambda_B}=R Z^2 (\frac{1}{2^2}-\frac{1}{\infty})=\frac14 RZ^2$.

The first line in the Paschen Series has $n=4$ :

$\frac{1}{\lambda_P}=RZ^2(\frac{1}{3^2}-\frac{1}{4^2})=RZ^2(\frac{1}{9}-\frac{1}{16})=\frac{7}{144}RZ^2$.

Therefore

$\frac{\lambda_B}{\lambda_P}=\frac{7}{144} \times \frac41 = \frac{7}{36}$.

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