Photons with energy 5, eV are incident on a c | vedclass

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(B) According to Einstein's photoelectric equation,the maximum kinetic energy of photoelectrons is given by:$KE_{\max} = E - \phi$For the first case,$

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(B) According to Einstein's photoelectric equation,the maximum kinetic energy of photoelectrons is given by:$KE_{\max} = E - \phi$For the first case,$E = 5\, eV$ and $KE_{\max} = 2\, eV$:$2 = 5 - \phi \implies \phi = 3\, eV$When photons of energy $E' = 6\, eV$ are incident,the new maximum kinetic energy is:$KE'_{\max} = E' - \phi = 6 - 3 = 3\, eV$For no photoelectrons to reach the anode $A,$ the potential of the anode relative to the cathode $(V_A - V_C)$ must be equal to the negative of the stopping potential $(V_s)$.The stopping potential $V_s$ is defined by $e V_s = KE'_{\max} = 3\, eV,$ so $V_s = 3\, V$.Since the anode must be at a negative potential relative to the cathode to stop the electrons,the potential difference $V_A - V_C = -3\, V$.

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