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(B) According to Einstein's photoelectric equation: $\frac{hc}{\lambda} - \phi = eV_0$, where $\phi$ is the work function.Using the data for $\lambda_

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$6.4 	imes 10^{-34}$

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(B) According to Einstein's photoelectric equation: $\frac{hc}{\lambda} - \phi = eV_0$, where $\phi$ is the work function.Using the data for $\lambda_1 = 0.3 \ \mu m$ and $\lambda_2 = 0.4 \ \mu m$:$1$) $\frac{hc}{0.3 	imes 10^{-6}} - \phi = 2e$$2$) $\frac{hc}{0.4 	imes 10^{-6}} - \phi = 1e$Subtracting equation $(2)$ from $(1)$:$hc \left( \frac{1}{0.3 	imes 10^{-6}} - \frac{1}{0.4 	imes 10^{-6}} ight) = 2e - e = e$$hc \left( \frac{0.4 - 0.3}{0.12 	imes 10^{-6}} ight) = e$$hc \left( \frac{0.1}{0.12 	imes 10^{-6}} ight) = e$$h = \frac{e 	imes 0.12 	imes 10^{-6}}{c 	imes 0.1} = \frac{1.6 	imes 10^{-19} 	imes 1.2 	imes 10^{-6}}{3 	imes 10^8}$$h = \frac{1.92 	imes 10^{-25}}{3 	imes 10^8} = 0.64 	imes 10^{-33} = 6.4 	imes 10^{-34} \ J \ s$.

$\lambda (\mu m)$	$V_0$ (Volt)
$0.3$	$2.0$
$0.4$	$1.0$
$0.5$	$0.4$

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