When photon of energy 4.25 eV strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy $T_{A}$ eV and de-Broglie wavelength $λ_{A}$ .The maximum kinetic energy of photoelectrons liberated from another metal B by photon of energy 4.70 eV is $T_{B} = (T_{A} - 1.50)$ eV. If the de-Broglie wavelength of these photoelectrons is $λ_{B} = 2 λ_{A}$ then

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the work function of B is 6.75 eV

$T_{B} = 2.75 eV$

$T_{A} = 2.00 eV$

the work function of A is 3.40 eV

answer is C.

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Detailed Solution

$K_{max} = E - W_{0} ∴ T_{A} = 4.25 - {(W_{0})}_{A} T_{B} = (T_{A} - 1.5) = 4.70 - {(W_{0})}_{B} Equation (i) and (ii) gives {(W_{0})}_{B} - {(W_{0})}_{A} = 1.95 eV De Broglie wave length λ = \frac{h}{\sqrt{2 mK}} \Rightarrow λ \propto \frac{1}{\sqrt{K}} \Rightarrow \frac{λ_{B}}{λ_{A}} = \sqrt{\frac{K_{A}}{K_{B}}} \Rightarrow 2 = \sqrt{\frac{T_{A}}{T_{B} - 1.5}} \Rightarrow T_{A} = 2 eV From equation (i) and (ii) W_{A} = 2.25 eV and W_{B} = 4.20 eV$