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

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$E_1 < 2E_2/3$

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(A) According to Einstein's photoelectric equation, the maximum kinetic energy $E_k$ is given by $E_k = \frac{hc}{\lambda} - \phi$, where $\phi$ is the work function of the metal.Since the metal plates are similar, $\phi$ is the same for both.For the first plate: $E_1 = \frac{hc}{\lambda_1} - \phi$For the second plate: $E_2 = \frac{hc}{\lambda_2} - \phi$Given $\lambda_1 = 1.5\lambda_2 = \frac{3}{2}\lambda_2$, we have $\frac{1}{\lambda_1} = \frac{2}{3\lambda_2}$.Substituting this into the expression for $E_1$: $E_1 = \frac{hc(2/3\lambda_2)} - \phi = \frac{2}{3}(\frac{hc}{\lambda_2}) - \phi$.Since $E_2 = \frac{hc}{\lambda_2} - \phi$, we have $\frac{hc}{\lambda_2} = E_2 + \phi$.Substituting this into the equation for $E_1$: $E_1 = \frac{2}{3}(E_2 + \phi) - \phi = \frac{2}{3}E_2 + \frac{2}{3}\phi - \phi = \frac{2}{3}E_2 - \frac{1}{3}\phi$.Since $\phi > 0$, it follows that $E_1 < \frac{2}{3}E_2$.

Photoelectrons are emitted from two similar metal plates when wavelengths $\lambda_1$ and $\lambda_2$ are incident on them $(\lambda_1 = 1.5\lambda_2)$. If the maximum kinetic energies of the emitted photoelectrons are $E_1$ and $E_2$ respectively, then:

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