A photon with a wavelength (λ) of 3.091 x 10–7 m strikes an atom of hydrogen. Determine the velocity (in meters/second) of an electron ejected from the excited state, n = 3.

Question

anonymous · Answer

Using this equation: 

$$E_{incident photon}=E_{binding}+E_{scattered electron}$$

And knowing:

$$E_{incident}=-h*c/\lambda$$

$$E_{scattered}=1/2*m*v^2$$

$$E_{binding}=-K*(1/n^2)$$

Being K=R*c*h (Rydberg constant, speed of light and Planck’s constant respectivily)

Then you can get the velocity substituting.

anonymous · Answer

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