As a result of collision with an electron of 20 eV energy, a hydrogen atom is excited to the higher energy state and the electron is scattered with a reduced velocity. Subsequecntly a photon with wavelength 1216 Å is emitted from the hydrogen atom . Determine the velocity of the electron after collision.

Explanation

Wavelength of the emitted photon,
`lambda =1216 "Å" =1216xx10^(-8) cm`
`therefore` The amount of energy gained by the hydrogen atom from the electron,
`E_2-E_1=hf=(hc)/(lambda)=(6.625xx10^(-27)xx3xx10^(10))/(1216xx10^(-8)) =1.634xx10^(-11) erg`
According to the questions, the initial kinetic energy of the electron `20 eV =20xx1.6xx10^(-12) erg=3.2xx10^(-11) erg `.
`therefore` The remaining kinetic energy of the electron after its collision with hydrogen atom is,
`1/2mv^2=(3.2-1.634)xx10^(-11)=1.566xx10^(-11) erg`
or, `v=sqrt((2xx1.566xx10^(-11))/(9.1xx10^(-28)))=1.855xx10^(8)cm*s^(-1)`