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(D) $1$. The stopping potential $V_s$ is related to the maximum kinetic energy $(K.E.)_{\max}$ by the equation: $(K.E.)_{\max} = e V_s$.Given $(K.E.)_

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$2 	ext{ V}, 3000 	ext{ Å}$

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(D) $1$. The stopping potential $V_s$ is related to the maximum kinetic energy $(K.E.)_{\max}$ by the equation: $(K.E.)_{\max} = e V_s$.Given $(K.E.)_{\max} = 3.2 	imes 10^{-19} 	ext{ J}$ and $e = 1.6 	imes 10^{-19} 	ext{ C}$.$V_s = \frac{(K.E.)_{\max}}{e} = \frac{3.2 	imes 10^{-19}}{1.6 	imes 10^{-19}} = 2 	ext{ V}$.$2$. The threshold wavelength $\lambda_0$ is related to the work function $\Phi$ by the equation: $\Phi = \frac{hc}{\lambda_0}$.Given $\Phi = 6.63 	imes 10^{-19} 	ext{ J}$,$h = 6.63 	imes 10^{-34} 	ext{ J} \cdot 	ext{s}$,and $c = 3 	imes 10^{8} 	ext{ m/s}$.$\lambda_0 = \frac{hc}{\Phi} = \frac{6.63 	imes 10^{-34} 	imes 3 	imes 10^{8}}{6.63 	imes 10^{-19}} = 3 	imes 10^{-7} 	ext{ m} = 3000 	imes 10^{-10} 	ext{ m} = 3000 	ext{ Å}$.Thus,the stopping potential is $2 	ext{ V}$ and the threshold wavelength is $3000 	ext{ Å}$. The correct option is $(D)$.

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