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Photons with energy 5 eV are incident on a cathode C in a photoelectric cell. The maximum energy of emitted photoelectrons is 2 eV. 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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+3 V

+4 V

-1 V

-3 V

answer is D.

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Detailed Solution

According Einstein's photoelectric equation kinetic energy of photoelectrons,

$\begin{matrix} K E_{\max} = E_{v} - ϕ \\ or, 2 = 5 - ϕ ∴ ϕ = 3 eV \end{matrix}$

$When E_{v} = 6 eV then$

$\begin{array}{l} K E_{\max} = 6 - 3 = 3 eV \\ or, e (V_{cathode} - V_{anode}) = 3 eV \\ or, V_{cathode} - V_{anode} = 3 V = - V_{stopping} \\ ∴ V_{stopping} = - 3 V \end{array}$

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