JEE Main · 2021 · 26 Feb 2021 · Shift 1 · Modern Physics

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λ_{1}

and

λ_{2}

are the wavelengths of the third member of Lyman and first member of the Paschen series respectively, then the value of

λ_{1} : λ_{2}

is :

(A) $1 : 3$
(B) $7 : 108$
(C) $1 : 9$
(D) $7 : 135$ ✓ Correct

Answer: (D)

Solution

For hydrogen spectrum,

\frac{1}{λ} = R (\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}})

. Third member of Lyman series:

n_{1} = 1, n_{2} = 4

, so

\frac{1}{λ _{1}} = R (1 - \frac{1}{16}) = \frac{15 R}{16}

. First member of Paschen series:

n_{1} = 3, n_{2} = 4

, so

\frac{1}{λ _{2}} = R (\frac{1}{9} - \frac{1}{16}) = R \frac{7}{144}

. Therefore

\frac{λ _{1}}{λ _{2}} = \frac{\frac{16}{15 R}}{\frac{144}{7 R}} = \frac{16 \times 7}{15 \times 144} = \frac{112}{2160} = \frac{7}{135}

JEE Main · 2021 · 26 Feb 2021 · Shift 1 · Single correct option

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