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An electron in the hydrogen atom jumps from excited state n to the ground state. The wavelength so emitted illuminates a photo sensitive metal having work function 2.75 eV. If the stopping potential of the photoelectron is 10 V, the value of n is ... |
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Answer» 3 `phi=-2.75eV` Total incident energy `E=phi+KE_(max)=12.75eV` `:.DeltaE=E_(n)-E_(1)` `12.75=(-13.6)/(n^(2))-(-(13.6)/(12))` `12.75=(-13.6)/(n^(2))+13.6` `(13.6)/(n^(2))=13.6-12.75` `=0.85` `:.(13.6)/(0.85)=n^(2)` `:.n^(2)=16` `:.n^(2)=4` |
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