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(D) According to Einstein's photoelectric equation: $eV_0 = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$,where $\lambda_0 = 5 	imes 10^{-10} \, m$ is t

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$< 0.5 V_0$

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(D) According to Einstein's photoelectric equation: $eV_0 = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$,where $\lambda_0 = 5 	imes 10^{-10} \, m$ is the threshold wavelength.For the first case,$\lambda_1 = 2 	imes 10^{-10} \, m$:$eV_0 = hc \left( \frac{1}{2 	imes 10^{-10}} - \frac{1}{5 	imes 10^{-10}} ight) = hc \left( \frac{5-2}{10 	imes 10^{-10}} ight) = \frac{3hc}{10 	imes 10^{-10}}$.For the second case,the wavelength is doubled: $\lambda_2 = 2 	imes \lambda_1 = 4 	imes 10^{-10} \, m$.Let the new stopping potential be $V_0'$.$eV_0' = hc \left( \frac{1}{4 	imes 10^{-10}} - \frac{1}{5 	imes 10^{-10}} ight) = hc \left( \frac{5-4}{20 	imes 10^{-10}} ight) = \frac{hc}{20 	imes 10^{-10}}$.Comparing $eV_0'$ and $eV_0$:$\frac{eV_0'}{eV_0} = \frac{hc / (20 	imes 10^{-10})}{3hc / (10 	imes 10^{-10})} = \frac{1}{20} 	imes \frac{10}{3} = \frac{1}{6}$.Therefore,$V_0' = \frac{V_0}{6}$,which is clearly less than $0.5 V_0$.

The threshold wavelength of a metal surface is $5 \times 10^{-10} \, m$. When it is illuminated by light of wavelength $2 \times 10^{-10} \, m$,the stopping potential is $V_0$. What will be the stopping potential if the wavelength of the incident light is doubled?

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