In Balmer series, wavelength of first line is ' λ₁ ' and in Brackett series wavelength of first line is ' λ₂ ' then (λ₁)/(λ₂) is (a) 0.162 (b) 0.124 (c) 0.138 (d) 0.188

Answer: (a). Balmer first line ( 3 → 2): (1)/(λ₁) = R(1/4-1/9) = 5R/36 Brackett first line ( 5 → 4): (1)/(λ₂) = R(1/16-1/25) = 9R/400 (λ₁)/(λ₂) = (36/5)/(400/9) = (324)/(2000) = 0.162 So the answer is (a) 0.162.

In Balmer series, wavelength of first line is ' λ₁ ' and in Brackett… | MHT CET 2019 (2 May, Shift 2)

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Hydrogen Spectrum & Spectral Series

In Balmer series, wavelength of first line is ' $\lambda_1$ ' and in Brackett series wavelength of first line is ' $\lambda_2$ ' then $\frac{\lambda_1}{\lambda_2}$ is

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Answer: (a)

Balmer first line ( $3 \to 2$ ): $\frac{1}{\lambda_1} = R\left(\frac{1}{4}-\frac{1}{9}\right) = \frac{5R}{36}$
Brackett first line ( $5 \to 4$ ): $\frac{1}{\lambda_2} = R\left(\frac{1}{16}-\frac{1}{25}\right) = \frac{9R}{400}$
$\frac{\lambda_1}{\lambda_2} = \frac{36/5}{400/9} = \frac{324}{2000} = 0.162$
So the answer is (a) 0.162.

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