What is the wavelength of the photon emitted during the transition from the orbit of n = 5 to that of n = 2 in hydrogen atom?

Given : 

n_i = 5, n_f = 2

To find : 

Wavelength of the photon emitted

Formulae :

i. \(\bar v\) = 109677 [\(\frac{1}{n_f^2}\) - \(\frac{1}{n_i^2}\)] cm^-1

ii. λ = \(\frac{1}{\bar v}\) 

Calculation : 

According to Rydberg equation [formula (i)],

\(\bar v\) = 109677 [\(\frac{1}{n_f^2}\) - \(\frac{1}{n_i^2}\)] cm^-1

Here,

n_i = 5, n_f = 2

∴ \(\bar v\) = 109677 [\(\frac{1}{2^2}\) - \(\frac{1}{5^2}\)] cm^-1

= 109677 [\(\frac{1}{4}\) - \(\frac{1}{25}\)] cm^-1

= 109677 [\(\frac{21}{100}\)] cm^-1

= 23032.17 cm^-1

From formula (ii),

λ = \(\frac{1}{\bar v}\)

= \(\frac{1}{23032.17}\)

= 4.34 x 10^-5 cm

∴ λ = 4.34 x 10^-5 x 10⁷nm

= 434 nm

Wavelength of the photon emitted is 434 nm.