Let K_1 be the maximum kinetic energy of phot | vedclass

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

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$K_1 < K_2$

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(C) 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 of the metal.For wavelength $\lambda_1$,$K_1 = \frac{hc}{\lambda_1} - \phi$.For wavelength $\lambda_2$,$K_2 = \frac{hc}{\lambda_2} - \phi$.Given $\lambda_1 = 2\lambda_2$,we substitute this into the expression for $K_1$:$K_1 = \frac{hc}{2\lambda_2} - \phi$.Since $\lambda_1 > \lambda_2$,the energy of the incident photon $\frac{hc}{\lambda_1}$ is less than $\frac{hc}{\lambda_2}$.Therefore,$K_1 = \frac{hc}{2\lambda_2} - \phi Thus,$K_1 < K_2$.

Let $K_1$ be the maximum kinetic energy of photoelectrons emitted by light of wavelength $\lambda_1$ and $K_2$ be the maximum kinetic energy corresponding to wavelength $\lambda_2$. If $\lambda_1 = 2\lambda_2$,then:

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