Question

What is the frequency and wavelength of a portion emitted during a transition from n=1 to n=2 state in the hydrogen atom

Answer 1

The formula for wave number is = 1.09677 $\times$ 10⁷($\frac{1}{{n_i}^2}-\frac{1}{{n_f}^2}$)m^-1

= 1.09677 $\times$10⁷($\frac{1}{1^2}-\frac{1}{2^2}$)
= 8.225 $\times$ 10⁶m^-1

Hence wavelength = $\frac{1}{wavenumber}=\frac{1}{8.225\times {10}^6}$=1.215 $\times$ 10^-7 m

Hence wavelength is 121.5 nm

​Now, frequency =$\frac{speedoflight}{wavelength}$=$\frac{3\times {10}^8}{1.215\times {10}^{-7}}$=2.469 $\times$10¹⁵Hz