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(B) According to Einstein's photoelectric equation,$E_K = E - \phi_0$,where $E_K$ is the kinetic energy,$E$ is the energy of the incident photon,and $

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$1.0$

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(B) According to Einstein's photoelectric equation,$E_K = E - \phi_0$,where $E_K$ is the kinetic energy,$E$ is the energy of the incident photon,and $\phi_0$ is the work function of the metal surface.From this,we have $E = E_K + \phi_0$ ... $(i)$.When the energy of the incident photon is increased by $20 \%$,the new energy $E'$ becomes $E' = E + 0.2E = 1.2E$.The new kinetic energy $E_K'$ is given by $E_K' = E' - \phi_0$,so $E' = E_K' + \phi_0$ ... (ii).Substituting $E' = 1.2E$ into equation (ii),we get $1.2E = E_K' + \phi_0$.From equation $(i)$,$E = 0.5 + \phi_0$. Substituting this into the modified equation (ii):$1.2(0.5 + \phi_0) = 0.8 + \phi_0$$0.6 + 1.2\phi_0 = 0.8 + \phi_0$$1.2\phi_0 - \phi_0 = 0.8 - 0.6$$0.2\phi_0 = 0.2$$\phi_0 = 1.0 \ eV$.

When the energy of incident radiation is increased by $20 \%$,the kinetic energy of the photoelectrons emitted from a metal surface increases from $0.5 \ eV$ to $0.8 \ eV$. The work function of the metal is (in $eV$)

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