Question

calculate the wave number of spectral line of shrtest wavelength appearing in the balmer series of H-spectrum.

Answer 1

HELLO THERE!

For an electron transition, the one having the shortest wavelength has the maximum energy. Hence, the shortest wavelength in Balmer series will be emitted when then electron will come to the first excited state (n₁ = 2) from infinity (n₂ = ∞).

We know,

[tex] \nu^{\prime} = R_{H}Z^{2}(\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}}) \\\\= 1.097\times10^{7}\times(\frac{1}{4} - 0}) \\\\= 2742500m^{-1} [/tex]

Where R is Rydberg Constant (= 1.097 x 10⁷ m⁻¹), Z is the atomic number (here, 1).

The wave number, thus obtained, is 2742500 m⁻¹.

HOPE THIS HELPS.

Thanks!