The value of Rydberg's constant is:

The correct answer is Option (1) → \( 1.03\times10^{7}\,\mathrm{m^{-1}} \)

The Rydberg constant R appears in the formula for the wavelengths of the hydrogen spectral lines, \( 1/\lambda = R\left(1/n_f^{2} - 1/n_i^{2}\right) \). Since it multiplies a pure number to give \( 1/\lambda \), its unit is \( \mathrm{m^{-1}} \), and its value is

    \( R \approx 1.1\times10^{7}\,\mathrm{m^{-1}} \)

Only Option (1) has this order of magnitude, \( 10^{7}\,\mathrm{m^{-1}} \), which is what decides the question.

A quick check: for the first Balmer line (n = 3 to n = 2), \( 1/\lambda = R(1/4 - 1/9) \approx 1.5\times10^{6}\,\mathrm{m^{-1}} \), so \( \lambda \approx 656\,\mathrm{nm} \), the familiar red line. A value of \( 10^{8} \) or a negative power of ten would give wavelengths far outside the visible region.