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Q.

Calculate the wave number of the highest energy transition in the Balmer series of Hydrogen atom [RH = 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  n2=∞ 

:.  in Balmer series of H-atom, n1=2 and n2=∞

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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