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(B) From Einstein's photoelectric equation: $4.25 = \phi_{A} + T_{A}$  --- $(1)$$4.70 = \phi_{B} + T_{B}$  --- $(2)$Given: $T_{B} = T_{A} - 1.50 \, eV

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The work function of $B$ is $4.20 \, eV$

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(B) From Einstein's photoelectric equation: $4.25 = \phi_{A} + T_{A}$  --- $(1)$$4.70 = \phi_{B} + T_{B}$  --- $(2)$Given: $T_{B} = T_{A} - 1.50 \, eV$De-Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$.Thus,$\frac{\lambda_{A}}{\lambda_{B}} = \sqrt{\frac{T_{B}}{T_{A}}}$.Given $\lambda_{B} = 2\lambda_{A}$,so $\frac{1}{2} = \sqrt{\frac{T_{B}}{T_{A}}} \Rightarrow \frac{1}{4} = \frac{T_{B}}{T_{A}} \Rightarrow T_{A} = 4T_{B}$.Substituting $T_{A} = 4T_{B}$ into the given relation: $T_{B} = 4T_{B} - 1.50 \Rightarrow 3T_{B} = 1.50 \Rightarrow T_{B} = 0.50 \, eV$.Then $T_{A} = 4 	imes 0.50 = 2.0 \, eV$.From $(1)$,$\phi_{A} = 4.25 - 2.0 = 2.25 \, eV$.From $(2)$,$\phi_{B} = 4.70 - 0.50 = 4.20 \, eV$.Therefore,the work function of $B$ is $4.20 \, eV$.

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