In a hydrogen spectrum, \(\lambda\) be the wavelength of first transition line of Lyman series. The wavelength difference will be "\(a\lambda\)" between the wavelength of \(3^{\text{rd}}\) transition line of Paschen series and that of \(2^{\text{nd}}\) transition line of Balmer Series where \(a=\) ______

JEE Main · 2022 · 26 Jul 2022 · Shift 1 · Modern Physics

In a hydrogen spectrum,

λ

be the wavelength of first transition line of Lyman series. The wavelength difference will be "

aλ

" between the wavelength of

3^{rd}

transition line of Paschen series and that of

2^{nd}

transition line of Balmer Series where

a =

______

Answer: 5

For the first line of Lyman series (n=2→1):

\frac{1}{λ} = R (1 - \frac{1}{4}) = \frac{3 R}{4} ⟹ λ = \frac{4}{3 R}

For the 3rd line of Paschen series (n=6→3):

\frac{1}{λ _{3}} = R (\frac{1}{9} - \frac{1}{36}) = R \cdot \frac{3}{36} = \frac{R}{12} ⟹ λ_{3} = \frac{12}{R}

For the 2nd line of Balmer series (n=4→2):

\frac{1}{λ _{2}} = R (\frac{1}{4} - \frac{1}{16}) = R \cdot \frac{3}{16} ⟹ λ_{2} = \frac{16}{3 R}

Wavelength difference:

aλ = λ_{3} - λ_{2} = \frac{12}{R} - \frac{16}{3 R} = \frac{36 - 16}{3 R} = \frac{20}{3 R}

Substituting

λ = \frac{4}{3 R}

a \cdot \frac{4}{3 R} = \frac{20}{3 R} ⟹ a = 5

JEE Main · 2022 · 26 Jul 2022 · Shift 1 · Integer answer