Question

7: find wavenumber, wavelength and frequency of Ha line of Hydrogen, assuming that the nucleus has infinite mass. Calculate the wavelength of the Balmer series limit.

Answer 1

Wave number is reciprocal of wavelength i.e. number of wavelength per centimeter. It is denoted by $\bar{v}$. H-alpha (H_α​) is a specific deep-red visible spectral line in the Balmer series; it occurs when a hydrogen electron falls from its third to second lowest energy level. For Hα​ - line in Balmer series of hydrogen spectrum $\bar{v}$ is given by $\bar{v}$ = R[$\frac{1}{2^2}$ - $\frac{1}{3^2}$ ] = $\frac{5R}{36}$ = 1.52 $\times$ 10⁶, where R is Rydberg constant.

Wavelength =$\frac{1}{wawe №}$ = 656.28 nm.

Frequency:

ΔE = hf = −13.6($\frac{1}{9} - \frac{1}{4}$) eV

ΔE = (13.6 $\times$ 5 / 36) $\times$ 1.6 $\times$ 10⁻¹⁹ J

f = $\frac{ΔE}{h}$ ​= $\frac{13.6 \times 5 \times 1.6 \times 10−19}{36 \times 6.6 \times 10−34}$ = 4.57 $\times$ 10¹⁴ Hz

For series limit of Balmer series,

p = 2 and n = ∞

$\frac{}{}$$\frac{1}{ λ}$ = R($\frac{1}{p^2}-\frac{1}{n^2}$) = R($\frac{1}{4} - \frac{1}{∞}$)

∴λ = $\frac{4}{R}$