For wavelength of visible radiation of Hydrogen spectrum, Balmer gave an equation as λ = (xm²)/(m² - 4), where m is the integer value. The value of x in terms of Rydberg's constant R is (a) R/4 (b) 4/R (c) 2R (d) 4R

Answer: (b). Balmer: (1)/(λ) = R(1/4 - 1/m²) = (R(m² - 4))/(4m²). λ = 4/R · (m²)/(m² - 4). Comparing with λ = (xm²)/(m² - 4) gives x = 4/R. So the answer is (b) 4/R.

For wavelength of visible radiation of Hydrogen spectrum, Balmer gave… | MHT CET 2026 (17 Apr, Shift 1)

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Hydrogen Spectrum & Spectral Series

For wavelength of visible radiation of Hydrogen spectrum, Balmer gave an equation as $\lambda = \frac{xm^2}{m^2 - 4}$ , where m is the integer value. The value of x in terms of Rydberg's constant R is

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Answer: (b)

Balmer: $\frac{1}{\lambda} = R\left(\frac{1}{4} - \frac{1}{m^2}\right) = \frac{R(m^2 - 4)}{4m^2}$ .
$\lambda = \frac{4}{R} \cdot \frac{m^2}{m^2 - 4}$ .
Comparing with $\lambda = \frac{xm^2}{m^2 - 4}$ gives $x = \frac{4}{R}$ .
So the answer is (b) $\frac{4}{R}$ .

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Hydrogen Spectrum & Spectral Series question from MHT CET 2026 (17 Apr, Shift 1)

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