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(C) According to Einstein's photoelectric equation: $E = \Phi + K_{max}$,where $E$ is the incident photon energy,$\Phi$ is the work function,and $K_{m

Vedclass · Accepted Answer

$-3.9$

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(C) According to Einstein's photoelectric equation: $E = \Phi + K_{max}$,where $E$ is the incident photon energy,$\Phi$ is the work function,and $K_{max}$ is the maximum kinetic energy of the emitted electrons.Given: $E = 6 \ eV$ and $\Phi = 2.1 \ eV$.Substituting the values: $6 \ eV = 2.1 \ eV + K_{max}$.Therefore,$K_{max} = 6 - 2.1 = 3.9 \ eV$.The stopping potential $V_0$ is related to the maximum kinetic energy by the relation $K_{max} = e V_0$.Thus,$V_0 = \frac{K_{max}}{e} = \frac{3.9 \ eV}{e} = 3.9 \ V$.Since the stopping potential is a negative potential applied to the anode to stop the most energetic electrons,it is represented as $-3.9 \ V$.

When a photon of energy $6 \ eV$ is incident on a metal surface with a work function of $2.1 \ eV$,what is the stopping potential of the emitted electrons in $V$?

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