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NCERT Class XI Chemistry Structure of Atom Solutions

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Question : 17 of 67
Marks: +1, -0
Calculate the wavenumber for the longest wavelength transition in the Balmer series of atomic hydrogen.
Solution:  
According to Rydberg equation,
1λ\frac{1}{\lambda} = vˉ(cm−1)\bar{v}(\mathrm{cm}^{-1}) = R(1n12−1n22)R\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) = 109678 (122−132)\left(\frac{1}{2^2} - \frac{1}{3^2}\right) = 109678 (14−19)\left(\frac{1}{4} - \frac{1}{9}\right)
= 109678 × 536 cm−1\frac{5}{36}\,\mathrm{cm}^{-1} = 1.5233 × 104 cm−110^{4}\,\mathrm{cm}^{-1}
Wavenumber of Balmer series = 1.523 × 106 m−110^{6}\,\mathrm{m}^{-1}
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