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(B) According to Einstein's photoelectric equation,the kinetic energy $K$ of emitted photoelectrons is given by $K = E - \phi_0$,where $E$ is the ener

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(B) According to Einstein's photoelectric equation,the kinetic energy $K$ of emitted photoelectrons is given by $K = E - \phi_0$,where $E$ is the energy of incident radiation and $\phi_0$ is the work function of the metal.Initially,the energy of incident radiation is $E$. Given $K_1 = 0.5\, eV$,we have:$0.5 = E - \phi_0$ ..... $(i)$When the energy of incident radiation is increased by $20\%$,the new energy becomes $E' = E + 0.2E = 1.2E$. The new kinetic energy is $K_2 = 0.8\, eV$. Thus:$0.8 = 1.2E - \phi_0$ ..... $(ii)$From equation $(i)$,we get $E = 0.5 + \phi_0$. Substituting this into equation $(ii)$:$0.8 = 1.2(0.5 + \phi_0) - \phi_0$$0.8 = 0.6 + 1.2\phi_0 - \phi_0$$0.8 - 0.6 = 0.2\phi_0$$0.2 = 0.2\phi_0$$\phi_0 = 1.0\, eV$.

When the energy of the 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 ............. $eV$.

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