Let K_1 be the maximum kinetic energy of phot | vedclass

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Question

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

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$K_1 < \frac{K_2}{2}$

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(C) According to Einstein's photoelectric equation,the maximum kinetic energy $K$ is given by $K = \frac{hc}{\lambda} - W$,where $W$ is the work function of the metal.For wavelength $\lambda_1$,$K_1 = \frac{hc}{\lambda_1} - W$  $(i)$For wavelength $\lambda_2$,$K_2 = \frac{hc}{\lambda_2} - W$  $(ii)$Given $\lambda_1 = 2\lambda_2$,substitute this into equation $(i)$:$K_1 = \frac{hc}{2\lambda_2} - W$$K_1 = \frac{1}{2} \left( \frac{hc}{\lambda_2} \right) - W$From equation $(ii)$,we know $\frac{hc}{\lambda_2} = K_2 + W$.Substituting this into the expression for $K_1$:$K_1 = \frac{1}{2} (K_2 + W) - W$$K_1 = \frac{K_2}{2} + \frac{W}{2} - W$$K_1 = \frac{K_2}{2} - \frac{W}{2}$Since $W > 0$,it follows that $K_1 < \frac{K_2}{2}$.

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

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