Question

Calculate the wavelength of light absorb when an electron jump from n=5 to n=6 state in hydrogen atom?

Answer 1

$E_n=-2.18×10^{-18}J(\frac{1}{n^2})$

$=-2.18×10^{-18}J(\frac{1}{n^2_f}-\frac{1}{n^2_i})$

$=-2.18×10^{-18}J(\frac{1}{6^2}-\frac{1}{5^2})$

=$-2.18×10^{-18}J(-\frac{11}{900})$

$=2.66×10^{-20}J$

$/\Delta E / =E_{photon}=hv =\frac{hc}{\lambda}$

Where h is Planck's constant,

c is the speed of light,

$\lambda$ is the wavelength of the incoming photon

Thus wavelength is

$\lambda=\frac{hc}{E_{photon}}$

$=\frac{(6.626×10^{-34}J.s)(2.998×10^8m/s)}{2.66×10^{-20}J}$

$=7.468×10^{-6}m$

$=74.68nm$