Ionization energy of hydrogen is 13.6 eV. If $h = 6 .6 \times 10^{- 34}$ J-s. the value of Rydberg constant will be of the order of

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$10^{- 7} m^{- 1}$

$10^{7} m^{- 1}$

$10^{10} m^{- 1}$

$10^{4} m^{- 1}$

answer is B.

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Detailed Solution

$E = - Rch \Rightarrow R = - \frac{E}{ch} = \frac{13 .6 \times 1 .6 \times 10^{- 19}}{3 \times 10^{8} \times 6 .6 \times 10^{- 34}} = 1 .098 \times 10^{7} per m$

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