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(A) The Einstein's photoelectric equation is given by $KE_{\max} = eV_S = \frac{hc}{\lambda} - \phi$.For $\lambda_1 = 280 \, nm$,the energy of the inc

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$1.3$

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(A) The Einstein's photoelectric equation is given by $KE_{\max} = eV_S = \frac{hc}{\lambda} - \phi$.For $\lambda_1 = 280 \, nm$,the energy of the incident photon is $E_1 = \frac{1240 \, eV \cdot nm}{280 \, nm} \approx 4.43 \, eV$.The stopping potential $V_{S1}$ is $V_{S1} = \frac{E_1 - \phi}{e} = 4.43 - 2.5 = 1.93 \, V$.For $\lambda_2 = 400 \, nm$,the energy of the incident photon is $E_2 = \frac{1240 \, eV \cdot nm}{400 \, nm} = 3.10 \, eV$.The stopping potential $V_{S2}$ is $V_{S2} = \frac{E_2 - \phi}{e} = 3.10 - 2.5 = 0.60 \, V$.The change in stopping potential is $\Delta V = V_{S1} - V_{S2} = 1.93 - 0.60 = 1.33 \, V \approx 1.3 \, V$.

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