Find the ratio between the wavelength of the most energetic' spectral lines in the Balmer and Paschen series of the hydrogen spectrum.

\(\frac1{\lambda}=R(\frac1{n_1^2}-\frac1{n_2^2})\) R → Rydburg constant

Balmer series

n₁ = 2

n₂ = \(\infty\)

for highest energy n₂ → \(\infty\) 

\(\frac1{\lambda_B}=R(\frac1{2^2}-\frac1{\infty})\) 

\(\frac1{\lambda_B}=R(\frac1{4})\) 

\(\frac1{\lambda_B}=R(\frac R{4})\) 

\({\lambda_B}=\frac4{R}\) 

for paschen series n₁ = 3, n₂ = \(\infty\)

\(\frac1{\lambda_p}=R(\frac{1}{(3^2)}-\frac1{(\infty^2)})\) 

\(\frac1{\lambda_p}=\frac{R}{p}\) 

\(\lambda_p=\frac{q}R,\) 

 \(\frac{\lambda_b}{\lambda_p}=\cfrac{\frac4R}{\frac{q}R}\) 

⇒  \(\frac4R\times\frac{R}q\)

⇒ 4 : q