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(A) According to Einstein's photoelectric equation,the energy of the incident photon is given by:$E = \phi + K_{\max}$$\frac{hc}{\lambda} = \frac{hc}{

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$4 \lambda$

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(A) According to Einstein's photoelectric equation,the energy of the incident photon is given by:$E = \phi + K_{\max}$$\frac{hc}{\lambda} = \frac{hc}{\lambda_0} + eV_s$For the first case,$\lambda_{	ext{incident}} = \lambda$ and $V_s = 3 	ext{ V}$:$\frac{hc}{\lambda} = \frac{hc}{\lambda_0} + 3e \quad \dots (i)$For the second case,$\lambda_{	ext{incident}} = 2\lambda$ and $V_s = 1 	ext{ V}$:$\frac{hc}{2\lambda} = \frac{hc}{\lambda_0} + 1e \quad \dots (ii)$Subtracting equation $(ii)$ from equation $(i)$:$\frac{hc}{\lambda} - \frac{hc}{2\lambda} = (\frac{hc}{\lambda_0} - \frac{hc}{\lambda_0}) + (3e - 1e)$$\frac{hc}{2\lambda} = 2e$$e = \frac{hc}{4\lambda}$Substitute the value of $e$ into equation $(ii)$:$\frac{hc}{2\lambda} = \frac{hc}{\lambda_0} + \frac{hc}{4\lambda}$$\frac{hc}{\lambda_0} = \frac{hc}{2\lambda} - \frac{hc}{4\lambda} = \frac{hc}{4\lambda}$$\lambda_0 = 4\lambda$

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