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Photons with energy 5eV are incident on a cathode C in a photoelectric cell The maximum energy of emitted photoelectrons is 2eV when photons of energy 6 eV are incident on C no photoelectrons will reach the anode A if the stopping potential of A relative to C is 

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 ev_{s}=hv-\phi =K.E_{max}

or\: 2ev=5ev-\phi

\therefore \phi =3ev

When electron is given 6ev by photon

its maximum kinetic energy will be  K.Emax = 6ev -3ev = 3ev

=> The stopping potential must be the minimum potential to stop this electron from reaching anode A. Then stopping potential of A relative

to C = -3ev

Posted by

avinash.dongre

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