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(B) According to Einstein's photoelectric equation, the maximum kinetic energy $K_{max}$ is given by $K_{max} = \frac{hc}{\lambda} - \phi$, where $\ph

Vedclass · Accepted Answer

$\frac{hc}{3\lambda}$

Vedclass · Answer

(B) According to Einstein's photoelectric equation, the maximum kinetic energy $K_{max}$ is given by $K_{max} = \frac{hc}{\lambda} - \phi$, where $\phi$ is the work function.For the first case with wavelength $\lambda$: $K_1 = \frac{hc}{\lambda} - \phi$.For the second case with wavelength $\lambda/3$: $K_2 = \frac{hc}{(\lambda/3)} - \phi = \frac{3hc}{\lambda} - \phi$.Given that $K_2 = 4K_1$, we substitute the expressions:$\frac{3hc}{\lambda} - \phi = 4(\frac{hc}{\lambda} - \phi)$.$\frac{3hc}{\lambda} - \phi = \frac{4hc}{\lambda} - 4\phi$.Rearranging the terms to solve for $\phi$:$4\phi - \phi = \frac{4hc}{\lambda} - \frac{3hc}{\lambda}$.$3\phi = \frac{hc}{\lambda}$.$\phi = \frac{hc}{3\lambda}$.

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