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Q.

Calculate the wave number of the highest energy transition in the Balmer series of Hydrogen atom [R_H= 109678 ${cm}^{- 1}$ ]

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a

27419.5 cm^–1

b

2741.95 cm^–1

c

274.195 cm^–1

d

274195.1cm^–1

answer is A.

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Detailed Solution

Maximum wavenumber $n_{2} = \infty$

:. in Balmer series of H-atom, $n_{1} = 2 and n_{2} = \infty$

large bar v = frac{1}{lambda } = {R_H}left( {frac{1}{{{2^2}}} - frac{1}{{{infty ^2}}}} right) = frac{{{R_H}}}{4}

large bar v = frac{{109678}}{4}c{m^{ - 1}} = 27419.5c{m^{ - 1}}

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