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When photons of $energy = 4 . 25 eV$ strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy, $T_{A}$ (expressed in eV ) and de-Broglie wavelength $λ_{A}$ . The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.70 eV is $T_{B} = (T_{A} - 1.50 e V)$ . If the de-Broglie wavelength of these photoelectrons is $λ_{B} = 2 λ_{A}$ , then

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the work function of A is 2.25 eV

the work function of B is 4.20 eV

$T_{A} = 2.00 e V$

$T_{B} = 2.75 e V$

answer is A, B, C.

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Detailed Solution

$\begin{array}{l} T_{A} = 4.25 - o A, λ_{A} = \frac{h}{\sqrt{2 m T_{A}}} \\ T_{B} = 4.70 - o B, λ_{B} = \frac{h}{\sqrt{2 m T_{B}}} \end{array}} λ_{B} = 2 λ_{A}$

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