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

Vedclass · Accepted Answer

$\frac{hc}{2 \lambda}$

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(B) According to Einstein's photoelectric equation,the maximum kinetic energy $K$ is given by $K = \frac{hc}{\lambda} - \phi$,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 $\frac{\lambda}{2}$: $K_2 = \frac{hc}{\lambda/2} - \phi = \frac{2hc}{\lambda} - \phi$.Given that $K_1 = \frac{1}{3} K_2$,we have $3K_1 = K_2$.Substituting the expressions: $3(\frac{hc}{\lambda} - \phi) = \frac{2hc}{\lambda} - \phi$.$\frac{3hc}{\lambda} - 3\phi = \frac{2hc}{\lambda} - \phi$.$\frac{3hc}{\lambda} - \frac{2hc}{\lambda} = 3\phi - \phi$.$\frac{hc}{\lambda} = 2\phi$.Therefore,$\phi = \frac{hc}{2\lambda}$.

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