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(A) According to Einstein's photoelectric equation:$eV = \frac{hc}{\lambda} - \Phi$ --- $(1)$where $V$ is the stopping potential,$\lambda$ is the inci

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$V - \frac{hc}{2e\lambda}$

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(A) According to Einstein's photoelectric equation:$eV = \frac{hc}{\lambda} - \Phi$ --- $(1)$where $V$ is the stopping potential,$\lambda$ is the incident wavelength,and $\Phi$ is the work function of the metal.When the incident wavelength is changed to $2\lambda$,let the new stopping potential be $V'$. The equation becomes:$eV' = \frac{hc}{2\lambda} - \Phi$ --- $(2)$Subtracting equation $(2)$ from equation $(1)$:$eV - eV' = \left( \frac{hc}{\lambda} - \Phi \right) - \left( \frac{hc}{2\lambda} - \Phi \right)$$e(V - V') = \frac{hc}{\lambda} - \frac{hc}{2\lambda} = \frac{hc}{2\lambda}$$V - V' = \frac{hc}{2e\lambda}$$V' = V - \frac{hc}{2e\lambda}$

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