Question

A ground state hydrogen atom absorbs a photon of light having a wavelength of 92.27nm. It then gives off a photon having a wavelength of 1817nm. What is the final state of the hydrogen atom? Values for physical constants can be found here. nf=?

Answer 1

dE = R*(1/nf^2 – 1/ni ^2)

Calculate dE:

Efinal - Einitial

Ef = hc/WLf

Ei = hc/WLi

where

h = Planck Constant = 6.626*10^-34 J s

c = speed of particle (i.e. light) = 3*10^8 m/s

Ef = hc/WLf = (6.626*10^-34)(3*10^8)/(1817*10^-9) = 1.094*10^-19  J/photon

Ei = hc/WLi = (6.626*10^-34)(3*10^8)/(92.27*10^-9) = 2.154*10^-18 J/photon

dE = ( 1.094*10^-19) - 2.154*10^-18 = 2.044*10^-18 J/photon

if ground state, then assum ni = 1

so, apply Rydbeg equation

dE = R*(1/nf^2 – 1/ni ^2)

R = -2.178*10^-18 J

-2.044*10^-18 = -2.178*10^-18*(1/nf^2 – 1/1^2)

(-2.044*10^-18) / (-2.178*10^-18) = 1/nf^2 -1

0.9384+1 = 1/nf

0.51 = 1/nf

1/0.51 = nf

nf = 2