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(B) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{\max}$ is given by $K_{\max} = eV_0 = hf - \phi_0$,where $V_0$ is th

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$2.4$

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(B) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{\max}$ is given by $K_{\max} = eV_0 = hf - \phi_0$,where $V_0$ is the stopping potential,$f$ is the frequency,and $\phi_0$ is the work function.For the initial case:$e(0.6) = hf - \phi_0 \quad \dots(i)$When the frequency is increased by $10 \%$,the new frequency $f' = f + 0.1f = 1.1f$. The new stopping potential is $V_0' = 0.9 \, V$.$e(0.9) = h(1.1f) - \phi_0 \quad \dots(ii)$From equation $(i)$,we have $hf = e(0.6) + \phi_0$. Substituting this into equation $(ii)$:$e(0.9) = 1.1(e(0.6) + \phi_0) - \phi_0$$e(0.9) = 1.1(e(0.6)) + 1.1\phi_0 - \phi_0$$e(0.9) = e(0.66) + 0.1\phi_0$$0.1\phi_0 = e(0.9 - 0.66)$$0.1\phi_0 = 0.24 \, eV$$\phi_0 = 2.4 \, eV$.

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