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 $K_{\max}$ is given by $K_{\max} = h
u - \phi_0$,where $h
u$ is the in

Vedclass · Accepted Answer

$-3$

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(B) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{\max}$ is given by $K_{\max} = h
u - \phi_0$,where $h
u$ is the incident photon energy and $\phi_0$ is the work function.For the first case: $2\, eV = 5\, eV - \phi_0$,which gives the work function $\phi_0 = 3\, eV$.For the second case: The incident energy is $6\, eV$. The maximum kinetic energy of the emitted photoelectrons is $K_{\max}' = 6\, eV - 3\, eV = 3\, eV$.To stop these photoelectrons from reaching the anode $A$,the potential of $A$ relative to $C$ $(V_{AC})$ must be negative and equal in magnitude to the maximum kinetic energy in electron volts. Thus,$V_{AC} = -3\, V$.

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