The wavelength of the first line in Balmer series in the hydrogen spectrum is ' λ '. What is the wavelength of the second line in the same series? (a) 20/27λ (b) 3/16λ (c) 5/36λ (d) 3/4λ

Answer: (a). First line ( 3 → 2): (1)/(λ) = 5R/36; second line ( 4 → 2): (1)/(λ') = R(1/4-1/16) = 3R/16 (λ')/(λ) = (5/36)/(3/16) = 80/108 = 20/27 So the answer is (a) 20/27λ.

The wavelength of the first line in Balmer series in the hydrogen… | MHT CET 2019 (2 May, Shift 1)

Q51MHT CET 2019 · 2 MAY · SHIFT 1Medium

Hydrogen Spectrum & Spectral Series

The wavelength of the first line in Balmer series in the hydrogen spectrum is ' $\lambda$ '. What is the wavelength of the second line in the same series?

Show answer and solution

Answer: (a)

First line ( $3 \to 2$ ): $\frac{1}{\lambda} = \frac{5R}{36}$ ; second line ( $4 \to 2$ ): $\frac{1}{\lambda'} = R\left(\frac{1}{4}-\frac{1}{16}\right) = \frac{3R}{16}$
$\frac{\lambda'}{\lambda} = \frac{5/36}{3/16} = \frac{80}{108} = \frac{20}{27}$
So the answer is (a) $\frac{20}{27}\lambda$ .

Report a mistake

Hydrogen Spectrum & Spectral Series question from MHT CET 2019 (2 May, Shift 1)

More questions on Hydrogen Spectrum & Spectral Series