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(A) According to Einstein's photoelectric equation,the energy of the incident photon $(E)$ is equal to the sum of the work function $(W_0)$ and the ma

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$2$

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(A) According to Einstein's photoelectric equation,the energy of the incident photon $(E)$ is equal to the sum of the work function $(W_0)$ and the maximum kinetic energy $(K_{max})$ of the emitted photoelectrons.$E = W_0 + K_{max}$Given that $K_{max} = eV_0$,where $V_0$ is the stopping potential,the equation becomes:$E = W_0 + eV_0$Substituting the given values ($E = 4 \ eV$ and $W_0 = 2 \ eV$):$4 \ eV = 2 \ eV + eV_0$$eV_0 = 4 \ eV - 2 \ eV$$eV_0 = 2 \ eV$Therefore,the stopping potential $V_0 = 2 \ V$.

$4 \ eV$ is the energy of the incident photon and the work function is $2 \ eV$. What is the stopping potential in $V$?

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