K_1 and K_2 are the maximum kinetic energies | vedclass

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

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$K_2 - 2K_1$

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(D) 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 wavelengths $\lambda_1$ and $\lambda_2$, we have:$K_1 = \frac{hc}{\lambda_1} - \phi$  --- $(1)$$K_2 = \frac{hc}{\lambda_2} - \phi$  --- $(2)$Given $\lambda_1 = 2\lambda_2$, substitute this into equation $(1)$:$K_1 = \frac{hc}{2\lambda_2} - \phi$Multiply by $2$:$2K_1 = \frac{hc}{\lambda_2} - 2\phi$  --- $(3)$Now, subtract equation $(3)$ from equation $(2)$:$K_2 - 2K_1 = (\frac{hc}{\lambda_2} - \phi) - (\frac{hc}{\lambda_2} - 2\phi)$$K_2 - 2K_1 = \phi$Therefore, the work function $\phi = K_2 - 2K_1$.

$K_1$ and $K_2$ are the maximum kinetic energies of photoelectrons emitted from a surface of a given material for light of wavelengths $\lambda_1$ and $\lambda_2$, respectively. If $\lambda_1 = 2\lambda_2$, then the work function of the material is given by:

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