A lithium ion with a single orbiting electron is subjected to ultraviolet radiation (λ = 3.711e-8 m). On interaction of this radiation with the lithium ion, the radiative energy is transferred to the electron, which is ejected from orbit as a result. As a free electron, it exhibits a velocity of 103 km/s. Given this information, determine the principal quantum number (n) of the electron in lithium prior to its interaction with the UV radiation.

Question

aaronq · Answer

try finding the energy of the photon:

$$E=\frac{ hc }{\lambda }$$

and subtract the KE the electron has (moving at 103 km/s) using it's velocity:

$$KE=\frac{ 1 }{ 2 }mv ^{2}$$

so, $$E _{photon}-KE _{electron}=E _{ionizing}$$


then use the rydberg formula knowing the energy required to ionize the electron

anonymous · Answer

plz ,@ aaronq I follow your steps but still some thing missing