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(D) According to Einstein's photoelectric equation,the maximum kinetic energy of an electron emitted from plate $1$ is $K_{max} = hv - \phi$.When the

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$hv$

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(D) According to Einstein's photoelectric equation,the maximum kinetic energy of an electron emitted from plate $1$ is $K_{max} = hv - \phi$.When the electron moves from plate $1$ to plate $2$,it is accelerated by the potential difference between the plates.The battery is connected such that plate $1$ is at a lower potential than plate $2$ (or the circuit configuration provides an accelerating potential). The potential difference $V$ across the capacitor is equal to the $emf$ of the battery,which is $V = \frac{\phi}{e}$.The work done by the electric field on the electron is $W = eV = e \cdot \frac{\phi}{e} = \phi$.The maximum kinetic energy of the electron reaching plate $2$ is the sum of its initial maximum kinetic energy and the work done by the electric field:$K_{final} = K_{max} + W = (hv - \phi) + \phi = hv$.Therefore,the correct option is $D$.

Two identical capacitors are arranged as shown. The work function of plate $1$ is $\phi \ eV$. The $emf$ of the battery is $\frac{\phi}{e}$. If the energy of the incident photon is $hv$,then what is the maximum kinetic energy of the $e^-$ reaching plate $2$?

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