In the hydrogen spectrum, λ be the wavelength of first transition… | JEE Main 2022 (26 Jul, Shift 1)

Q39JEE Main 2022 · 26 JUL · SHIFT 1Difficult

Hydrogen Spectrum & Spectral Series

In the hydrogen spectrum, $\lambda$ be the wavelength of first transition line of Lyman series. The wavelength difference will be "a $\lambda$ '' between the wavelength of $3$ rd transition line of Paschen series and that of $2$ nd transition line of Balmer series where $a =$ ___________.

Numerical answer type. Work it out, then open the answer.

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Answer: 5

Lyman first line ( $2 \to 1$ ): $\frac{1}{\lambda} = \frac{3R}{4} \Rightarrow \lambda = \frac{4}{3R}$
Paschen 3rd line ( $6 \to 3$ ): $\frac{1}{\lambda_P} = R\left(\frac{1}{9}-\frac{1}{36}\right) = \frac{R}{12} \Rightarrow \lambda_P = \frac{12}{R}$
Balmer 2nd line ( $4 \to 2$ ): $\lambda_B = \frac{16}{3R}$
$\lambda_P-\lambda_B = \frac{20}{3R} = 5\times\frac{4}{3R} = 5\lambda$
Answer: $a = 5$

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