Find the velocity of photoelectrons liberated by electromagnetic radiation of wavelength `lambda=18.0nm` from stationary `He^(+)` ions in the ground state.

Find the velocity of photoelectrons liberated by electromagnetic radia

1.	Find the velocity of photoelectrons liberated by electromagnetic radiation of wavelength `lambda=18.0nm` from stationary `He^(+)` ions in the ground state.
Answer» By conservation of enery `(1)/(2)mv^(2)=(2pi ħ c)/(lambda)-E_(b)` where `E_(b)=4 ħ R` is the binding energy of the electron in the ground state of `He^(+)`. (Recoil of `He^(++)` nucleus is neglected). Then `v=sqrt((2)/(m)((2pi ħc)/(lambda)-E_(b)))` Substitution gives `v= 2.25xx10^(6)m//s`

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Find the velocity of photoelectrons liberated by electromagnetic radiation of wavelength `lambda=18.0nm` from stationary `He^(+)` ions in the ground state.

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