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(D) Using Einstein's photoelectric equation: $K_{max} = E - \Phi$,where $\Phi$ is the work function.For metal $A$: $T_A = 4.25 - \Phi_A$ ... $(i)$For

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All of the above

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(D) Using Einstein's photoelectric equation: $K_{max} = E - \Phi$,where $\Phi$ is the work function.For metal $A$: $T_A = 4.25 - \Phi_A$ ... $(i)$For metal $B$: $T_B = T_A - 1.50 = 4.70 - \Phi_B$ ... (ii)From $(i)$,$\Phi_A = 4.25 - T_A$. From (ii),$\Phi_B = 4.70 - (T_A - 1.50) = 6.20 - T_A$.The de-Broglie wavelength is given by $\lambda = \frac{h}{\sqrt{2mK}}$,so $\lambda \propto \frac{1}{\sqrt{K}}$.Given $\lambda_B = 2\lambda_A$,we have $\frac{\lambda_B}{\lambda_A} = \sqrt{\frac{T_A}{T_B}} = 2$.Squaring both sides: $\frac{T_A}{T_A - 1.50} = 4$.$T_A = 4T_A - 6.00 \Rightarrow 3T_A = 6.00 \Rightarrow T_A = 2.00\, eV$.Substituting $T_A = 2.00\, eV$ into $(i)$: $\Phi_A = 4.25 - 2.00 = 2.25\, eV$.Substituting $T_A = 2.00\, eV$ into (ii): $T_B = 2.00 - 1.50 = 0.50\, eV$. Then $\Phi_B = 4.70 - 0.50 = 4.20\, eV$.Thus,all statements are correct.

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